Exercise 3.4 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.4$
Question 1.
Expand the following :
(i) $(2 x+3 y+4 z)^{2}$
(ii) $(-\mathrm{p}+2 \mathrm{q}+3 \mathrm{r})^{2}$
(iii) $(2 \mathrm{p}+3)(2 \mathrm{p}-4)(2 \mathrm{p}-5)$
(iv) $(3 a+1)(3 a-2)(3 a+4)$
Solution:
(i) $(2 x+3 y+4 z)^{2}$
$(a+b+c)^{2} \equiv a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$
$\therefore(2 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z})^{2}=(2 \mathrm{x})^{2}+(3 \mathrm{y})^{2}+(4 \mathrm{z})^{2}+2(2 \mathrm{x})(3 \mathrm{y})+2(3 \mathrm{y})(4 \mathrm{z})+2(4 \mathrm{z})(2 \mathrm{x})$ $=4 x^{2}+9 y^{2}+16 z^{2}+12 x y+24 y z+16 z x$
(ii) $(-\mathrm{p}+2 \mathrm{q}+3 \mathrm{r})^{2}$
$(a+b+c)^{2} \equiv a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$
$(-p+2 q+3 r)^{2}=(-p)^{2}+(2 q)^{2}+(3 r)^{2}+2(-p)(2 q)+2(2 q)(3 r)+2(-p)(3 r)$
$=\mathrm{p}^{2}+4 \mathrm{q}^{2}+9 \mathrm{r}^{2}-4 \mathrm{pq}+12 \mathrm{qr}-6 \mathrm{rp}$
(iii) $(2 p+3)(2 p-4)(2 p-5)$
$(x+a)(x+b)(x+c) \equiv x^{3}+(a+b+c) x^{2}+(a b+b e+c a) x+a b c$
$(2 p+3)(2 p-4)(2 p-5)=(2 p)^{3}+(3-4-5)(2 p)^{2}+\left[(3 \times-4)^{2}+(-4 \times-5)+(-5 \times 3)\right] 2 p+3$ $x-4 \times-5$
$=8 \mathrm{p}^{3}+(-6)\left(4 \mathrm{p}^{2}\right)+[-12+20+(-15)] 2 \mathrm{p}+60$
$=8 \mathrm{p}^{3}-24 \mathrm{p}^{2}+(-7) 2 \mathrm{p}+60$
$(2 p+3)(2 p-4)(2 p-5)=8 p^{3}-24 p^{2}-14 p+60$

$\begin{aligned}
&\text { (iv) }(3 a+1)(3 a-2)(3 a+4) \\
&(x+a)(x+b)(x+c) \equiv x^{3}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c \\
&(3 a+1)(3 a-2)(3 a+4)=(3 a)^{3}+(1-2+4)(3 a)^{2}+[1 \times(-2)+(-2 \times 4)+4 \times 1](3 a)+1 \times \\
&-2 \times 4 \\
&=27 a^{3}+3\left(9 a^{2}\right)+(-2-8+4) 3 a-8 \\
&=27 a^{3}+27 a^{2}-8 a-8
\end{aligned}$

Question 2.
Using algebraic identity, find the coefficients of $x 2, x$ and constant term without actual expansion.
(i) $(x+5)(x+6)(x+7)$
(ii) $(2 x+3)(2 x-5)(2 x-6)$
Solution:
(i) $(x+5)(x+6)(x+7)$
$(x+a)(x+b)(x+c) \equiv x^{3}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c$
Co-efficient of $\mathrm{x}^{2}=\mathrm{a}+\mathrm{b}+\mathrm{c}=5+6+7=18$
Co-efficient of $\mathrm{x}^{2}=\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=(5 \times 6)+(6 \times 7)+(7 \times 5)$
$=30+42+35=107$
Constant term $=a b c=5 \times 6 \times 7$
Co-efficient of constant term $=210$

(ii) $(2 x+3)(2 x-5)(2 x-6)$
$\therefore$ Co-efficient of $\mathrm{x}^{2}=4(\mathrm{a}+\mathrm{b}+\mathrm{c})=4(3+(-5)+(-6))$
$=4 \times(-8)=-32$
Co-efficient of $x=2(a b+b c+c a)$
$=2[3 \times(-5)+(-5)(-6)+(-6)(3)]$
$=2[-15+30-18]=2 \times(-3)=-6$
Constant term $=a b c=3 \times(-5) \times(-6)=90$
 

Question $3 .$
If $(x+a)(x+b)(x+c)=x^{3}+14 x^{2}+59 x+70$, find the value of
(i) $a+b+c$
(ii) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
(iii) $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}$
(iv) $\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}$
Solution:
$(x+a)(x+b)(x+c)=x^{3}+14 x^{2}+59 x+70$
$(x+a)(x+b)(x+c) \equiv x^{3}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c$
(i) Comparing (1) \& (2)
We get, $a+b+c=14$
(ii) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{b c+a c+a b}{a b c}=\frac{59}{70}$
(iii) $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$=14^{2}-2(59)=196-118=78$
(iv) $\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}=\frac{78}{70}=\frac{39}{35}$

Question 4.

Expand

(i) $(3 a-4 b)^{3}$
(ii) $\left(x+\frac{1}{y}\right)^{3}$
Solution:
(i) $(3 a-4 b)^{3}$ We know that
$\begin{aligned}
&(x-y)^{3}=x^{3}-3 x^{2} y+3 x y^{2}-y^{3} \\
&(3 a-4 b)^{3}=(3 a)^{3}-3(3 a)^{2}(4 b)+3(3 a)(4 b)^{2}-(4 b)^{3} \\
&=27 a^{3}-108 a^{2} b+144 a b^{2}-64 b^{3}
\end{aligned}$
(ii) $\left(x+\frac{1}{y}\right)^{3}$
$\begin{aligned}
&(\mathrm{x}+\mathrm{y})^{3} \equiv \mathrm{x}^{3}+3 \mathrm{x}^{2} \mathrm{y}+3 \mathrm{xy}^{2}+\mathrm{y}^{3} \\
&\left(x+\frac{1}{y}\right)^{3}=x^{3}+\frac{3 x^{2}}{y}+\frac{3 x}{y^{2}}+\frac{1}{y^{3}}
\end{aligned}$

Question $5 .$
Evaluate the following by using identities:
(i) $98^{3}$
(ii) $1001^{3}$
Solution:
(i) $98^{2}=(100-2)^{3}$
$(a-b)^{3} \equiv a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$
$98^{3}=(100-2)^{3}=100^{3}-3 \times 100^{2} \times+3 \times 100 \times 2^{2}-2^{3}$
$=1000000-3 \times 10000 \times 2+300 \times 4-8$
$=1000000-60000+1200-8=1001200-60008=941192$

(ii) $1001^{3}=(1000+1)^{3}$
$\begin{aligned}
&(a+b)^{3} \equiv a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
&(1000+1)^{3}=1000^{3}+3\left(1000^{2}\right) \times 1+3 \times 1000 \times 1^{2}+1^{3} \\
&=1000,000,000+3,000,000+3000+1=1,003,003,001
\end{aligned}$

Question $6 .$
If $(x+y+z)=9$ and $(x y+y z+z x)=26$ then find the value of $x^{2}+y^{2}+z^{2}$
Solution:
$\begin{aligned}
&(x+y+z)=9 \text { and }(x y+y z+z x)=26 \\
&x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
&=9^{2}-2 \times 26=81-52=29
\end{aligned}$

Question $7 .$
Find $27 a^{3}+64 b^{3}$, if $3 a+4 b=10$ and $a b=2$.
Solution:
$3 a+4 b=10, a b=2$
$\begin{aligned}
&(3 a+4 b)^{3}=(3 a)^{3}+3(3 a)^{2}(4 b)+3(3 a)(4 b)^{2}+(4 b)^{3} \\
&\left(27 a^{3}+64 b^{3}\right)=(3 a+4 b)^{3}-3(3 a)(4 b)(3 a+4 b)
\end{aligned}$

$\begin{aligned}
&\because x^{3}+y^{3}=(x+y)^{3}-3 x y-(x+y) \\
&=10^{3}-36 a b(10)=1000-36 \times 2 \times 10 \\
&=1000-720=280
\end{aligned}$

Question 8.
Find $x^{3}-y^{3}$, if $x-y=5$ and $x y=14$.
Solution:
$\begin{aligned}
&x-y=5, x y=14 \\
&x^{3}-y^{3}=(x-y)^{3}+3 x y(x-y)=5^{3}+3 \times 14 \times 5 \\
&=125+210=335
\end{aligned}$

Question 9
. If $\mathrm{a}+\frac{1}{a}=6$, then find the value of $\mathrm{a}^{3}+\frac{1}{a^{3}}$
Solution:
$\begin{aligned}
&a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b) \\
&a^{3}+\left(\frac{1}{a}\right)^{3}=\left(a+\frac{1}{a}\right)^{3}-3 a \times \frac{1}{a}\left(a+\frac{1}{a}\right) \\
&a^{3}+\frac{1}{a^{3}}=6^{3}-3 \times 6=216-18=198
\end{aligned}$

Question $10 .$
If $x^{2}+\frac{1}{x^{2}}=23$, then find the value of $x+\frac{1}{x}$ and $x^{3}+\frac{1}{x^{3}}$

Solution:
$\begin{aligned}
\left(x+\frac{1}{x}\right)^{2} &=x^{2}+2 x \times \frac{1}{\not x}+\frac{1}{x^{2}} \\
\left(x+\frac{1}{x}\right)^{2} &=x^{2}+2+\frac{1}{x^{2}}=\left(x^{2}+\frac{1}{x^{2}}\right)+2 \\
\left(x+\frac{1}{x}\right)^{2} &=23+2=25 \\
\left(x+\frac{1}{x}\right) &=\sqrt{25}=5 \\
x^{3}+\frac{1}{x^{3}} &=\left(x+\frac{1}{x}\right)^{3}-3 x \times \frac{1}{x}\left[x+\frac{1}{x}\right] \\
&=5^{3}-3(5)=125-15=110
\end{aligned}$
 

Question 11

 If $\left(\mathrm{y}-\frac{1}{y}\right)^{3}=27$, then find the value of $\mathrm{y}^{3}-\frac{1}{y^{3}}$
Solution:
$\begin{aligned}
&\left(y-\frac{1}{y}\right)^{3}=27(\text { Given }) \\
&y^{3}-\frac{1}{y^{3}}=\left(y-\frac{1}{y}\right)^{3}+3 y+\frac{1}{y}\left(y-\frac{1}{y}\right)
\end{aligned}$

$\begin{aligned}
{\left[\because x^{3}-y^{3}\right.} & \equiv(x-y)^{3}+3 x y(x-y) \\
\because\left(y-\frac{1}{y}\right)^{3} &=27 ; y-\frac{1}{y}=\sqrt[3]{27}=3 \\
&=27+3 \times 3=27+9=36
\end{aligned}$

Question 12
- Simplify : (i) $(2 a+36+4 c)\left(4 a+9 b^{2}+16 c^{2}-6 a b-12 b c-12 b c-8 c a\right)$
(ii) $(x-2 y+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 x z\right)$
Solution:
(i) $(2 a+36+4 c)\left(4 a^{2}+9 b^{2}+16 c^{2}-6 a b-12 b c-12 b c-8 c a\right)$
We know that
$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3} \times 3 a b c$
$\begin{aligned}
&\therefore(2 a+36+4 c)\left(4 a^{2}+9 b^{2}+16 c^{2}-6 a b-\right. \\
&=(2 a)^{3}+(3 b)^{3}+(4 c)^{3}-3 \times 2 a \times 36 \times 4 c \\
&=8 a^{3}+27 b^{3}+64 c^{3}-72 a b c
\end{aligned}$
(ii) $(x-2, y+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 x z\right)$
$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$
$\begin{aligned}
&\therefore(x-2 y+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 x z\right) \\
&=x^{3}+(-2 y)^{3}+(3 z)^{3}-3 \times x \times(-2 y)(3 z) \\
&=x^{3}-8 y^{3}+27 z^{3}+18 x y z
\end{aligned}$

Question 13
. By using identity evaluate the following:
(i) $7^{3}-10^{3}+3^{3}$
(ii) $1+\frac{1}{8}-\frac{27}{8}$
Solution:
(i) $7^{3}-10^{3}+3^{3}$
$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$
If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$
$\therefore 7-10+3=0$
$\begin{aligned}
&\Rightarrow 7^{3}-10^{3}+3^{3}=3 \times 7 \times-10 \times 3 \\
&=9 \times-70=-630
\end{aligned}$
(ii) $1+\frac{1}{8}-\frac{27}{8}$
$1+\frac{1}{8}-\frac{27}{8}=1^{3}+\left(\frac{1}{2}\right)^{3}+\left(\frac{-3}{2}\right)^{3}$ Here $1+\frac{1}{2}-\frac{3}{2}=\frac{2+1-3}{2}=\frac{0}{2}=0$
$\therefore 1+\frac{1}{8}-\frac{27}{8}=1^{3}+\left(\frac{1}{2}\right)^{3}+\left(\frac{-3}{2}\right)^{3}=3 \times 1 \times \frac{1}{2} \times \frac{-3}{2}=\frac{-9}{4}$

Question $14 .$
If $2 x-3 y-4 z=0$, then find $8 x^{3}-27 y^{3}-64 z^{3}$.
Solution:
$\begin{aligned}
&\text { If } 2 x-3 y-4 z=0 \text { then } 8 x^{3}-27 y^{3}-64 z^{3}=? \\
&\text { If } x+y+z=0 \text { then } x^{3}+y^{3}+z^{3}=3 x y z \\
&8 x^{3}-21 y^{3}-64 z^{3}=(2 x)^{3}+(-3 y)^{3}+(-4 z)^{3} \\
&=3 \times 2 x \times-3 y \times-4 z=72 x y z
\end{aligned}$