NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities - Ganita Manjari
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NCERT Solutions for Chapter 4 Exploring Algebraic Identities Class 9 Maths
Page No. 42
Think and Reflect
Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Answer
Consider four consecutive numbers \(n,\ n+1,\ n+2,\ n+3\). Their squares are \(n^2,\ (n+1)^2,\ (n+2)^2,\ (n+3)^2\).
Consider the difference between the sum of the first and fourth squares and the sum of the second and third squares:
\[ [n^2+(n+3)^2]-[(n+1)^2+(n+2)^2] \]
\[ = [n^2+n^2+6n+9]-[n^2+2n+1+n^2+4n+4] \]
\[ =2n^2+6n+9-(2n^2+6n+5)=4 \]
Therefore, for any four consecutive square numbers, the sum of the first and fourth squares exceeds the sum of the second and third squares by 4.
For example, 1,4,9,16:
(1+16) - (4+9) = 17-13 = 4.
Think and Reflect
1. What can you say about a and b if \((a+b)^2 < a^2+b^2\)?
Answer
Using the identity
\[ (a+b)^2=a^2+2ab+b^2, \]
we get
\[ (a+b)^2-(a^2+b^2)=2ab. \]
Since \((a+b)^2<a^2+b^2\), we have \(2ab<0\).
Therefore,
\[ ab<0. \]
Hence, a and b have opposite signs.
2. What can you say about a and b if \((a+b)^2 > a^2+b^2\)?
Answer
Since
\[ (a+b)^2-(a^2+b^2)=2ab, \]
and \((a+b)^2>a^2+b^2\), we have \(2ab>0\).
Therefore,
\[ ab>0. \]
Hence, a and b have the same sign.
3. When will \((a+b)^2\) be equal to \(a^2+b^2\)?
Answer
We require
\[ (a+b)^2=a^2+b^2. \]
Using the identity,
\[ a^2+2ab+b^2=a^2+b^2. \]
Thus,
\[ 2ab=0, \]
so
\[ ab=0. \]
Therefore, \((a+b)^2\) is equal to \(a^2+b^2\) when a = 0 or b = 0.
Exercise Set 4.1
1. Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), expand the following:
(i) \((7x+4y)^2\)
Answer
\[ (7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2 \]
\[ =49x^2+56xy+16y^2. \]
Therefore, \((7x+4y)^2=49x^2+56xy+16y^2\).
(ii) \(\left(\frac{7}{5}x+\frac{3}{2}y\right)^2\)
Answer
\[ \left(\frac{7}{5}x+\frac{3}{2}y\right)^2\]
\[=\frac{49}{25}x^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+\frac{9}{4}y^2 \]
\[ =\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2 \]
(iii) \((2.5p+1.5q)^2\)
Answer
\[ (2.5p+1.5q)^2 \]
\[=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2 \]
\[ =6.25p^2+7.5pq+2.25q^2 \]
(iv) \(\left(\frac{3}{4}s+8t\right)^2\)
Answer
\[ \left(\frac{3}{4}s+8t\right)^2 \]
\[ =\frac{9}{16}s^2+12st+64t^2 \]
(v) \(\left(x+\frac{1}{2y}\right)^2\)
Answer
\[ \left(x+\frac{1}{2y}\right)^2 \]
\[ =x^2+2\left(x\right)\left(\frac{1}{2y}\right)+\frac{1}{4y^2} \]
\[ =x^2+\frac{x}{y}+\frac{1}{4y^2} \]
(vi) \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Answer
\[ \left(\frac{1}{x}+\frac{1}{y}\right)^2\]
\[ =\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2} \]
2. Using the same identity, find the values of the following:
(i) \((64)^2\)
Answer
\[ 64^2=(60+4)^2 \]
\[ =60^2+2(60)(4)+4^2\]
\[ =3600+480+16=4096\]
(ii) \((105)^2\)
Answer
\[ 105^2=(100+5)^2 \]
\[ =10000+1000+25=11025\]
(iii) \((205)^2\)
Answer
\[ 205^2=(200+5)^2\]
\[=40000+2000+25=42025\]
Think and Reflect
What if we replace b by – b in \((a+b)^2 = a^2+2ab+b^2\)?
Answer
Replacing \(b\) by \(-b\), we get
\[ (a+(-b))^2=a^2+2a(-b)+(-b)^2 \]
Therefore,
\[ (a-b)^2=a^2-2ab+b^2. \]
Thus, the new identity is
\[ \boxed{(a-b)^2=a^2-2ab+b^2}. \]
Exercise Set 4.2
1. Factor completely:
(i) \(9x^2+24xy+16y^2\)
Answer
\[ 9x^2+24xy+16y^2 \]
\[=(3x)^2+2(3x)(4y)+(4y)^2 \]
\[ \boxed{=(3x+4y)^2} \]
(ii) \(4s^2+20st+25t^2\)
Answer
\[ 4s^2+20st+25t^2\]
\[ =(2s)^2+2(2s)(5t)+(5t)^2 \]
\[ \boxed{=(2s+5t)^2} \]
(iii) \(49x^2+28xy+4y^2\)
Answer
\[ 49x^2+28xy+4y^2\]
\[ =(7x)^2+2(7x)(2y)+(2y)^2 \]
\[ \boxed{=(7x+2y)^2} \]
(iv) \(64p^2+\frac{32}{3}pq+\frac{4}{9}q^2\)
Answer
\[ 64p^2+\frac{32}{3}pq+\frac{4}{9}q^2\]
\[=(8p)^2+2(8p)\left(\frac{2}{3}q\right)+\left(\frac{2}{3}q\right)^2 \]
\[ \boxed{=\left(8p+\frac{2}{3}q\right)^2} \]
(v) \(3a^2+4ab+\frac{4}{3}b^2\)
Answer
Taking \(\frac{1}{3}\) common,
\[ 3a^2+4ab+\frac{4}{3}b^2
\[=\frac{1}{3}(9a^2+12ab+4b^2) \]
\[ =\frac{1}{3}(3a+2b)^2. \]
\[ \boxed{=\frac{1}{3}(3a+2b)^2} \]
(vi) \(\frac{9}{5}s^2+6sv+5v^2\)
Answer
Taking \(\frac{1}{5}\) common,
\[ \frac{9}{5}s^2+6sv+5v^2 \]
\[=\frac{1}{5}(9s^2+30sv+25v^2) \]
\[ =\frac{1}{5}(3s+5v)^2. \]
\[ \boxed{=\frac{1}{5}(3s+5v)^2} \]
2. Find the values of the following using the identity \((a-b)^2=a^2-2ab+b^2\).
(i) \((79)^2\)
Answer
\[ 79^2=(80-1)^2 \]
\[=6400-160+1=6241 \]
(ii) \((193)^2\)
Answer
\[ 193^2=(200-7)^2\]
\[=40000-2800+49=37249 \]
(iii) \((299)^2\)
Answer
\[ 299^2=(300-1)^2\]
\[ =90000-600+1=89401 \]
Think and Reflect
Label the squares and rectangles in Fig. 4.4 so that it represents the identity \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\).
Answer
The large square has side \(a+b+c\), so its area is \((a+b+c)^2\).
The regions in Fig. 4.4 should be labelled as follows:
- Square of side \(a\): \(a^2\)
- Square of side \(b\): \(b^2\)
- Square of side \(c\): \(c^2\)
- Two rectangles of dimensions \(a\times b\): \(ab,\ ab\)
- Two rectangles of dimensions \(b\times c\): \(bc,\ bc\)
- Two rectangles of dimensions \(c\times a\): \(ca,\ ca\)
Hence, the total area is
\[ a^2+b^2+c^2+2ab+2bc+2ca. \]
Therefore,
\[ \boxed{(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca}. \]
Exercise Set 4.3
1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) \(117^2\)
Answer
Using \((a+b)^2=a^2+2ab+b^2\),
\[ 117^2=(100+17)^2 \]
\[=10000+3400+289 =13689. \]
(ii) \(78^2\)
Answer
Using \((a-b)^2=a^2-2ab+b^2\),
\[ 78^2=(80-2)^2=6400-320+4=6084. \]
(iii) \(198^2\)
Answer
\[ 198^2=(200-2)^2=40000-800+4=39204 \]
(iv) \(214^2\)
Answer
\[ 214^2=(200+14)^2=40000+5600+196=45796. \]
(v) \(1104^2\)
Answer
\[ 1104^2=(1100+4)^2 =1210000+8800+16 =1218816. \]
(vi) \(1120^2\)
Answer
\[ 1120^2=(1100+20)^2 =1210000+44000+400 =1254400. \]
2. Factor using suitable identities:
(i) \(16y^2-24y+9\)
Answer
\[ 16y^2-24y+9=(4y-3)^2. \]
(ii) \(\frac{9}{4}s^2+6st+4t^2\)
Answer
\[ \frac{9}{4}s^2+6st+4t^2 =\left(\frac{3}{2}s+2t\right)^2. \]
(iii) \(\frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+3nk+2mn+9n^2\)
Answer
\[ \frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+3nk+2mn+9n^2 \]
\[ =\left(\frac{m}{3}+\frac{k}{2}+3n\right)^2. \]
(iv) \(\frac{p^2}{16}-2+\frac{16}{p^2}\)
Answer
\[ \frac{p^2}{16}-2+\frac{16}{p^2} =\left(\frac{p}{4}-\frac{4}{p}\right)^2. \]
(v) \(9a^2+4b^2+c^2-12ab+6ac-4bc\)
Answer
\[ 9a^2+4b^2+c^2-12ab+6ac-4bc \]
\[ =(3a-2b+c)^2. \]
3. Expand the following using the identity \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\):
(i) \((p+3q+7r)^2\)
Answer
\[ (p+3q+7r)^2 =p^2+9q^2+49r^2+6pq+42qr+14pr. \]
(ii) \((3x-2y+4z)^2\)
Answer
\[ (3x-2y+4z)^2 =9x^2+4y^2+16z^2-12xy-16yz+24xz. \]
4. Is this an identity?
\[ (a+b-c)^2+(a-b+c)^2+(a-b-c)^2=2a^2+2b^2+2c^2. \]
Answer
Expand each term:
\[ (a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca \]
\[ (a-b+c)^2=a^2+b^2+c^2-2ab-2bc+2ca \]
\[ (a-b-c)^2=a^2+b^2+c^2-2ab+2bc+2ca. \]
Adding them,
\[ (a+b-c)^2+(a-b+c)^2+(a-b-c)^2 \]
\[ =3a^2+3b^2+3c^2-2ab-2bc+2ca. \]
This is not equal to \(2a^2+2b^2+2c^2\) for all values of \(a,b,c\).
Therefore, it is not an identity.
Think and Reflect
1. Try to evaluate the following using a suitable identity:
(i) \(35^2\) (ii) \(65^2\) (iii) \(85^2\) (iv) \(105^2\)
Do you observe any interesting pattern?
Answer
\[ 35^2=(30+5)^2=900+300+25=1225 \]
\[ 65^2=(60+5)^2=3600+600+25=4225 \]
\[ 85^2=(80+5)^2=6400+800+25=7225 \]
\[ 105^2=(100+5)^2=10000+1000+25=11025. \]
Thus,
\[ 35^2=1225,\quad65^2=4225,\quad85^2=7225,\quad105^2=11025. \]
The interesting pattern is that the square of every number ending in 5 ends in 25.
2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Answer
The identity represented by Fig. 4.6 is
\[ \boxed{(a+b+c)^2+(a+b-c)^2+(a-b+c)^2+(a-b-c)^2 =4a^2+4b^2+4c^2}. \]
It follows by adding the four expansions. All the cross-product terms cancel, leaving \(4a^2+4b^2+4c^2\).
Think and Reflect
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Answer
No. If \(7x\) is split as \(2x+5x\), then the corresponding constant terms would have to multiply to \(12\).
The possible factor pairs of \(12\) are \(1,12\), \(2,6\), and \(3,4\).
Only \(3+4=7\). Therefore, the required rectangular arrangement is possible only with \(3x+4x\), giving
\[ (x+3)(x+4)=x^2+7x+12. \]
For the split \(2x+5x\), the corresponding product of constants would be \(2\times5=10\), not \(12\).
Think and Reflect
1. Figure out the product of x + 2 and x + 3 using algebra tiles.
Answer
Using the distributive property,
\[ (x+2)(x+3)=x^2+3x+2x+6 \]
\[ \boxed{(x+2)(x+3)=x^2+5x+6}. \]
Thus, the algebra tiles consist of one \(x^2\)-tile, five \(x\)-tiles and six unit tiles.
2. Lay out algebra tiles for \(x^2 + 11x + 30\) in such a way that you will see its factors.
Answer
We need two numbers whose sum is \(11\) and product is \(30\).
\[ 5+6=11,\qquad5\times6=30. \]
Therefore, arrange the \(11x\)-tiles as \(5x+6x\) and the \(30\) unit tiles as a \(5\times6\) array.
The rectangular dimensions are \(x+5\) and \(x+6\).
\[ \boxed{x^2+11x+30=(x+5)(x+6)}. \]
Think and Reflect
We have seen that \((x+3)(x+4)=x^2+7x+12\). Also \((x+6)(x+7)=x^2+13x+42\). Generalise the pattern to get an expression for \((x+a)(x+b)\).
Answer
Using the distributive property,
\[ (x+a)(x+b)=x(x+b)+a(x+b) \]
\[ =x^2+bx+ax+ab \]
\[ \boxed{(x+a)(x+b)=x^2+(a+b)x+ab}. \]
Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1, as shown in Fig. 4.8. What can you say about its area (2x + 3) (3x + 1)?
Answer
\[ (2x+3)(3x+1) =6x^2+2x+9x+3 \]
\[ \boxed{=6x^2+11x+3}. \]
Fill in the blanks with the appropriate expressions to make the equation true.
\[ (px+a)(qx+b)=(\_\_\_\_\_)x^2+(\_\_\_\_\_)x+\_\_\_\_\_. \]
Answer
\[ (px+a)(qx+b) =pq x^2+pbx+aqx+ab \]
\[ \boxed{(px+a)(qx+b)=pqx^2+(pb+aq)x+ab}. \]
For \(2x+3\) and \(3x+1\), \(p=2,\ q=3,\ a=3,\ b=1\).
\[ pq=6,\qquad pb+aq=2(1)+3(3)=11,\qquad ab=3. \]
Hence,
\[ \boxed{(2x+3)(3x+1)=6x^2+11x+3}. \]
Exercise Set 4.4
1. Fill in the blanks to complete the following identities:
(i) \(s^2-11s+24=(\_\_\_\_\_\_\_\_) (\_\_\_\_\_\_\_\_)\)
Answer
We need two numbers whose sum is \(-11\) and product is \(24\): \(-3\) and \(-8\).
\[ \boxed{s^2-11s+24=(s-3)(s-8)}. \]
(ii) \((\_\_\_\_\_\_\_\_) (x+1)=3x^2-4x-7\)
Answer
\[ 3x^2-4x-7=3x^2+3x-7x-7 \]
\[ =3x(x+1)-7(x+1) \]
\[ \boxed{=(3x-7)(x+1)}. \]
(iii) \(10x^2-11x-6=(2x-\_\_\_) (\_\_\_+2)\)
Answer
We have
\[ (2x-3)(5x+2) =10x^2+4x-15x-6 \]
\[ =10x^2-11x-6. \]
Therefore,
\[ \boxed{10x^2-11x-6=(2x-3)(5x+2)}. \]
(iv) \(6x^2+7x+2=(\_\_\_\_\_\_\_\_\_\_\_\_) (\_\_\_\_\_\_\_\_\_\_\_\_)\)
Answer
Split \(7x\) as \(4x+3x\):
\[ 6x^2+7x+2=6x^2+4x+3x+2 \]
\[ =2x(3x+2)+(3x+2) \]
\[ \boxed{=(2x+1)(3x+2)}. \]
2. Select and use the identity that will help you to find the following products without multiplying directly:
(i) \((41)^2\)
Answer
\[ 41^2=(40+1)^2=1600+80+1=1681. \]
Answer: \(1681\)
(ii) \((27)^2\)
Answer
\[ 27^2=(30-3)^2=900-180+9=729. \]
Answer: \(729\)
(iii) \((23\times17)\)
Answer
\[ 23\times17=(20+3)(20-3) \]
\[ =20^2-3^2=400-9=391. \]
Answer: \(391\)
(iv) \((135)^2\)
Answer
\[ 135^2=(130+5)^2 =16900+1300+25 =18225. \]
Answer: \(18225\)
(v) \((97)^2\)
Answer
\[ 97^2=(100-3)^2=10000-600+9=9409. \]
Answer: \(9409\)
(vi) \((18\times29)\)
Answer
\[ 18\times29=(23.5-5.5)(23.5+5.5) \]
\[ =23.5^2-5.5^2 =552.25-30.25 =522. \]
Answer: \(522\)
(vii) \((34\times43)\)
Answer
\[ 34\times43=(38.5-4.5)(38.5+4.5) \]
\[ =38.5^2-4.5^2 =1482.25-20.25 =1462. \]
Answer: \(1462\)
(viii) \((205)^2\)
Answer
\[ 205^2=(200+5)^2 =40000+2000+25 =42025. \]
Answer: \(42025\)
Think and Reflect
James and Reshma were talking about algebraic identities they learnt in school.
James: \((a-b)^2(a+b)=(a^2-2ab+b^2)(a+b)\)
Reshma: I have a different idea. \((a-b)^2(a+b)=(a-b)[(a-b)(a+b)]\)
\(=(a-b)(a^2-b^2)\)
I will find this product to get the answer.
According to you, who is correct and why?
Try to combine more such identities and find new results.
Answer
Both James and Reshma are correct.
James uses the identity
\[ (a-b)^2=a^2-2ab+b^2 \]
and then multiplies the result by \((a+b)\).
Reshma first uses the identity
\[ (a-b)(a+b)=a^2-b^2. \]
Therefore,
\[ (a-b)^2(a+b) =(a-b)[(a-b)(a+b)] \]
\[ =(a-b)(a^2-b^2). \]
Expanding this gives
\[ (a-b)(a^2-b^2) =a^3-a^2b-ab^2+b^3. \]
Hence, both approaches give the same result:
\[ \boxed{(a-b)^2(a+b)=a^3-a^2b-ab^2+b^3}. \]
Reshma's method is more efficient because it combines two known identities before expanding.
Exercise Set 4.5
1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) \(\frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}\)
Answer
\[ 3p^2-3pq-18q^2=3(p^2-pq-6q^2) \]
\[ =3(p-3q)(p+2q). \]
Also,
\[ p^2+3pq-10q^2=(p+5q)(p-2q). \]
Therefore,
\[ \boxed{\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}}. \]
(ii) \(\frac{n^3-3n^2m+3nm^2-m^3}{5m^2-10mn+5n^2}\)
Answer
The numerator is
\[ n^3-3n^2m+3nm^2-m^3=(n-m)^3. \]
The denominator is
\[ 5m^2-10mn+5n^2=5(n-m)^2. \]
Therefore,
\[ \frac{(n-m)^3}{5(n-m)^2} =\boxed{\frac{n-m}{5}}. \]
(iii) \(\frac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}\)
Answer
Using \(a=w,\ b=-v,\ c=x\) in
\[ a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca), \]
we get
\[ w^3-v^3+x^3+3wvx =(w-v+x)(w^2+v^2+x^2-w(-v)-(-v)x-xw). \]
This gives the denominator exactly.
Hence,
\[ \boxed{w-v+x}. \]
(iv) \(\frac{4y^2-20yz+25z^2}{25z^2-4y^2}\)
Answer
\[ 4y^2-20yz+25z^2=(2y-5z)^2=(5z-2y)^2. \]
Also,
\[ 25z^2-4y^2=(5z-2y)(5z+2y). \]
Therefore,
\[ \frac{(5z-2y)^2}{(5z-2y)(5z+2y)} =\boxed{\frac{5z-2y}{5z+2y}}. \]
(v) \(\frac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}\)
Answer
Factor each expression:
\[ x^2+x-6=(x+3)(x-2), \]
\[ x^2-7x+12=(x-3)(x-4), \]
\[ x^2-6x+8=(x-2)(x-4), \]
\[ x^2-9=(x-3)(x+3). \]
Therefore,
\[ \frac{(x+3)(x-2)(x-3)(x-4)} {(x-2)(x-4)(x-3)(x+3)} =\boxed{1}. \]
(vi) \(\frac{p^4-16}{p^2-4p+4}\)
Answer
\[ p^4-16=(p^2-4)(p^2+4) \]
\[ =(p-2)(p+2)(p^2+4). \]
Also,
\[ p^2-4p+4=(p-2)^2. \]
Therefore,
\[ \frac{(p-2)(p+2)(p^2+4)}{(p-2)^2} =\boxed{\frac{(p+2)(p^2+4)}{p-2}}. \]
Think and Reflect
Try to simplify the following rational expression:
\[ \frac{36s^2-12st+t^2}{t^2+2ts-48s^2} =\frac{(6s-t)^2}{(\_\_\_+\_\_\_)(\_\_\_+\_\_\_)}. \]
(Hint: Factor \(t^2+2ts-48s^2\) and simplify the rational expressions assuming that \(t^2+2ts-48s^2\neq0\)).
Answer
The numerator is
\[ 36s^2-12st+t^2=(6s-t)^2=(t-6s)^2. \]
For the denominator,
\[ t^2+2ts-48s^2 =(t+8s)(t-6s). \]
Therefore,
\[ \frac{(t-6s)^2}{(t+8s)(t-6s)} =\boxed{\frac{t-6s}{t+8s}}. \]
End-of-Chapter Exercises
1. Use suitable identities to find the following products:
(i) \((-3x+4)^2\)
Answer
\[ (-3x+4)^2=9x^2-24x+16. \]
(ii) \((2s+7)(2s-7)\)
Answer
Using \((a+b)(a-b)=a^2-b^2\),
\[ (2s+7)(2s-7)=4s^2-49. \]
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
Answer
\[ \left(p^2+\frac12\right)\left(p^2-\frac12\right) =p^4-\frac14. \]
(iv) \((2n+7)(2n-7)\)
Answer
\[ (2n+7)(2n-7)=4n^2-49. \]
(v) \((s-2t)(s^2+2st+4t^2)\)
Answer
Using
\[ (a-b)(a^2+ab+b^2)=a^3-b^3, \]
with \(a=s\) and \(b=2t\),
\[ (s-2t)(s^2+2st+4t^2) =s^3-(2t)^3 \]
\[ \boxed{=s^3-8t^3}. \]
(vi) \(\left(\frac{1}{2r}-4r\right)^2\)
Answer
\[ \left(\frac{1}{2r}-4r\right)^2 =\frac{1}{4r^2}-4+16r^2. \]
(vii) \((-3m+4k-l)^2\)
Answer
Using \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\),
\[ (-3m+4k-l)^2 =9m^2+16k^2+l^2-24mk-8kl+6ml. \]
(viii) \(\left(x-\frac{1}{3}y\right)^3\)
Answer
Using
\[ (a-b)^3=a^3-3a^2b+3ab^2-b^3, \]
\[ \left(x-\frac13y\right)^3 =x^3-x^2y+\frac13xy^2-\frac1{27}y^3. \]
(ix) \(\left(\frac{7}{2}k-\frac{2}{3}m\right)^3\)
Answer
Using \((a-b)^3=a^3-3a^2b+3ab^2-b^3\),
\[ \left(\frac72k-\frac23m\right)^3 =\frac{343}{8}k^3-\frac{49}{2}k^2m+\frac{14}{3}km^2-\frac{8}{27}m^3. \]
2. Find the values using suitable identities:
(i) \(17\times21\)
Answer
\[ 17\times21=(19-2)(19+2)=19^2-2^2 =361-4=357. \]
Answer: \(357\)
(ii) \(104\times96\)
Answer
\[ 104\times96=(100+4)(100-4) =10000-16 =9984. \]
Answer: \(9984\)
(iii) \(24\times16\)
Answer
\[ 24\times16=(20+4)(20-4) =400-16 =384. \]
Answer: \(384\)
(iv) \(147^3\)
Answer
\[ 147^3=(150-3)^3 \]
\[ =150^3-3(150)^2(3)+3(150)(3)^2-3^3 \]
\[ =3375000-202500+4050-27 \]
\[ \boxed{=3176523}. \]
(v) \(199^3\)
Answer
\[ 199^3=(200-1)^3 \]
\[ =8000000-120000+600-1 \]
\[ \boxed{=7880599}. \]
(vi) \(127^3\)
Answer
\[ 127^3=(125+2)^3 \]
\[ =125^3+3(125)^2(2)+3(125)(2)^2+2^3 \]
\[ =1953125+93750+1500+8 \]
\[ \boxed{=2048383}. \]
(vii) \((-107)^3\)
Answer
\[ (-107)^3=-(107)^3 \]
\[ =-(100+7)^3 \]
\[ =-\left(1000000+210000+14700+343\right) \]
\[ \boxed{=-1225043}. \]
(viii) \((-299)^3\)
Answer
\[ (-299)^3=-(300-1)^3 \]
\[ =-\left(27000000-270000+900-1\right) \]
\[ \boxed{=-26730899}. \]
3. Factor the following algebraic expressions:
(i) \(4y^2+1+\frac{1}{16y^2}\)
Answer
\[ 4y^2+1+\frac{1}{16y^2} =\left(2y+\frac{1}{4y}\right)^2. \]
(ii) \(9m^2-\frac{1}{25n^2}\)
Answer
Using \(a^2-b^2=(a+b)(a-b)\),
\[ 9m^2-\frac{1}{25n^2} =\left(3m-\frac{1}{5n}\right) \left(3m+\frac{1}{5n}\right). \]
(iii) \(27b^3-\frac{1}{64b^3}\)
Answer
Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\), take
\[ a=3b,\qquad b=\frac{1}{4b}. \]
Therefore,
\[ 27b^3-\frac{1}{64b^3} =\left(3b-\frac{1}{4b}\right) \left(9b^2+\frac34+\frac{1}{16b^2}\right). \]
(iv) \(x^2+\frac{5x}{6}+\frac{1}{6}\)
Answer
We need two numbers whose sum is \(\frac56\) and product is \(\frac16\).
\[ \frac12+\frac13=\frac56,\qquad \frac12\times\frac13=\frac16. \]
Hence,
\[ \boxed{x^2+\frac{5x}{6}+\frac16 =\left(x+\frac12\right)\left(x+\frac13\right)}. \]
(v) \(27u^3-\frac{1}{125}-\frac{27u^2}{5}+\frac{9u}{25}\)
Answer
Rearranging the terms,
\[ 27u^3-\frac{27u^2}{5}+\frac{9u}{25}-\frac1{125}. \]
This is of the form \(a^3-3a^2b+3ab^2-b^3\), with
\[ a=3u,\qquad b=\frac15. \]
Therefore,
\[ \boxed{27u^3-\frac{1}{125}-\frac{27u^2}{5}+\frac{9u}{25} =\left(3u-\frac15\right)^3}. \]
(vi) \(64y^3+\frac{1}{125}z^3\)
Answer
Using \(a^3+b^3=(a+b)(a^2-ab+b^2)\), take
\[ a=4y,\qquad b=\frac15z. \]
Thus,
\[ \boxed{ 64y^3+\frac1{125}z^3 =\left(4y+\frac15z\right) \left(16y^2-\frac45yz+\frac1{25}z^2\right) }. \]
(vii) \(p^3+27q^3+r^3-9pqr\)
Answer
Using
\[ a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca), \]
take \(a=p,\ b=3q,\ c=r\).
Therefore,
\[ \boxed{ p^3+27q^3+r^3-9pqr =(p+3q+r)(p^2+9q^2+r^2-3pq-3qr-pr) }. \]
(viii) \(9m^2-12m+4\)
Answer
\[ 9m^2-12m+4=(3m-2)^2. \]
(ix) \(9x^3-\frac{8}{3}y^3+\frac{z^3}{3}+6xyz\)
Answer
Take \(a=3x,\ b=-2y,\ c=z\). Then
\[ a^3+b^3+c^3-3abc =27x^3-8y^3+z^3+18xyz. \]
Therefore, the given expression is \(\frac13\) times this identity:
\[ \boxed{ 9x^3-\frac83y^3+\frac13z^3+6xyz =\frac13(3x-2y+z) (9x^2+4y^2+z^2+6xy+2yz-3xz) }. \]
(x) \(4x^2+9y^2+36z^2+12xz+36yz+24xy\)
Answer
\[ 4x^2+9y^2+36z^2+12xz+36yz+24xy \]
\[ \boxed{=(2x+3y+6z)^2}. \]
(xi) \(27u^3-\frac{1}{216}-\frac{9u^2}{2}+\frac{u}{4}\)
Answer
Take
\[ a=3u,\qquad b=\frac16. \]
Using \((a-b)^3=a^3-3a^2b+3ab^2-b^3\),
\[ \boxed{ 27u^3-\frac1{216}-\frac{9u^2}{2}+\frac{u}{4} =\left(3u-\frac16\right)^3 }. \]
4. Simplify the following:
(i) \(\frac{4x^2+4x+1}{4x^2-1}\)
Answer
\[ 4x^2+4x+1=(2x+1)^2 \]
and
\[ 4x^2-1=(2x-1)(2x+1). \]
Therefore,
\[ \frac{(2x+1)^2}{(2x-1)(2x+1)} =\boxed{\frac{2x+1}{2x-1}}. \]
(ii) \(\frac{9(3a^3-24b^3)}{9a^2-36b^2}\)
Answer
\[ 3a^3-24b^3=3(a^3-8b^3) \]
\[ =3(a-2b)(a^2+2ab+4b^2). \]
Also,
\[ 9a^2-36b^2=9(a^2-4b^2) =9(a-2b)(a+2b). \]
Hence,
\[ \frac{9\cdot3(a-2b)(a^2+2ab+4b^2)} {9(a-2b)(a+2b)} \]
\[ \boxed{=\frac{3(a^2+2ab+4b^2)}{a+2b}}. \]
(iii) \(\frac{s^3+125t^3}{s^2-2st-35t^2}\)
Answer
\[ s^3+125t^3 =s^3+(5t)^3 =(s+5t)(s^2-5st+25t^2). \]
Also,
\[ s^2-2st-35t^2=(s-7t)(s+5t). \]
Therefore,
\[ \boxed{ \frac{s^3+125t^3}{s^2-2st-35t^2} =\frac{s^2-5st+25t^2}{s-7t} }. \]
5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) \(25a^2-30ab+9b^2\)
Answer
\[ 25a^2-30ab+9b^2=(5a-3b)^2. \]
Therefore, possible expressions for both length and breadth are
\[ \boxed{5a-3b\text{ units and }5a-3b\text{ units}}. \]
(ii) \(36s^2-49t^2\)
Answer
\[ 36s^2-49t^2=(6s-7t)(6s+7t). \]
Therefore, possible expressions are
\[ \boxed{6s-7t\text{ units and }6s+7t\text{ units}}. \]
6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) \(6a^2-24b^2\)
Answer
\[ 6a^2-24b^2=6(a^2-4b^2) \]
\[ =6(a-2b)(a+2b). \]
Hence, possible dimensions are
\[ \boxed{6,\quad a-2b,\quad a+2b}. \]
(ii) \(3ps^2-15ps+12p\)
Answer
Taking \(3p\) common,
\[ 3ps^2-15ps+12p =3p(s^2-5s+4) \]
\[ =3p(s-1)(s-4). \]
Hence, possible dimensions are
\[ \boxed{3p,\quad s-1,\quad s-4}. \]
7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Answer
The path is created around all four sides, so the side of the outer square is
\[ 40+2s. \]
Area of outer square:
\[ (40+2s)^2. \]
Area of playground:
\[ 40^2=1600. \]
Therefore, area of the path is
\[ (40+2s)^2-1600 \]
\[ =1600+160s+4s^2-1600 \]
\[ \boxed{4s^2+160s\text{ square metres}}. \]
8. If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Answer
Let the number be \(x\). Then
\[ x+\frac1x=\frac{10}{3}. \]
Multiplying by \(3x\),
\[ 3x^2+3=10x. \]
\[ 3x^2-10x+3=0. \]
Factorising,
\[ (3x-1)(x-3)=0. \]
Therefore,
\[ 3x-1=0\quad\text{or}\quad x-3=0. \]
Hence,
\[ \boxed{x=\frac13\text{ or }x=3}. \]
9. A rectangular pool has area \(2x^2+7x+3\) square hastas. If its width is \(2x+1\) hastas, find its length. Hasta was a unit used to measure length.
Answer
The area is
\[ 2x^2+7x+3. \]
Factorise:
\[ 2x^2+7x+3 =2x^2+6x+x+3 \]
\[ =2x(x+3)+(x+3) \]
\[ =(2x+1)(x+3). \]
Since width is \(2x+1\) hastas, the length is
\[ \boxed{x+3\text{ hastas}}. \]
*10. If both \(x-2\) and \(x-\frac{1}{2}\) are factors of \(px^2+5x+r\), show that \(p=r\).
Answer
Since \(x-2\) and \(x-\frac12\) are factors,
\[ px^2+5x+r=p\left(x-2\right)\left(x-\frac12\right). \]
Now,
\[ \left(x-2\right)\left(x-\frac12\right) =x^2-\frac52x+1. \]
Therefore,
\[ px^2+5x+r =px^2-\frac{5p}{2}x+p. \]
Comparing the coefficient of \(x\),
\[ -\frac{5p}{2}=5. \]
Thus,
\[ p=-2. \]
Comparing the constant terms gives
\[ r=p. \]
Hence,
\[ \boxed{p=r}. \]
*11. If \(a+b+c=5\) and \(ab+bc+ca=10\), then prove that \(a^3+b^3+c^3-3abc=-25\).
Answer
We use the identity
\[ a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \]
First,
\[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca). \]
Therefore,
\[ 25=a^2+b^2+c^2+20, \]
so
\[ a^2+b^2+c^2=5. \]
Hence,
\[ a^3+b^3+c^3-3abc =5(5-10) \]
\[ \boxed{=-25}. \]
*12. By factoring the expression, check that \(n^3-n\) is always divisible by 6 for all natural numbers \(n\). Give reasons.
Answer
Factorise:
\[ n^3-n=n(n^2-1) \]
\[ =n(n-1)(n+1). \]
Thus, \(n^3-n\) is the product of three consecutive natural numbers.
Among any three consecutive integers:
- one is divisible by \(3\);
- at least one is even and therefore divisible by \(2\).
Hence their product is divisible by \(2\times3=6\).
Therefore,
\[ \boxed{n^3-n\text{ is always divisible by }6}. \]
*13. Find the value of
(i) \(x^3+y^3-12xy+64\), when \(x+y=-4\)
Answer
Using
\[ x^3+y^3=(x+y)^3-3xy(x+y), \]
and \(x+y=-4\),
\[ x^3+y^3=(-4)^3-3xy(-4) \]
\[ =-64+12xy. \]
Therefore,
\[ x^3+y^3-12xy+64 =(-64+12xy)-12xy+64 \]
\[ \boxed{=0}. \]
(ii) \(x^3-8y^3-36xy-216\), when \(x=2y+6\)
Answer
Using \(x=2y+6\), we have
\[ x-2y=6. \]
Now use
\[ x^3-(2y)^3=(x-2y)(x^2+2xy+4y^2). \]
Thus,
\[ x^3-8y^3-36xy-216 \]
\[ =6(x^2+2xy+4y^2)-36xy-216. \]
Since \(x=2y+6\),
\[ x^2+2xy+4y^2 =(2y+6)^2+2(2y+6)y+4y^2 \]
\[ =12y^2+36y+36. \]
Therefore,
\[ 6(12y^2+36y+36)-36xy-216. \]
Substituting \(x=2y+6\),
\[ =72y^2+216y+216-36(2y+6)y-216 \]
\[ =72y^2+216y+216-72y^2-216y-216 \]
\[ \boxed{=0}. \]
