Exercise 3.3 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Expand
(i) $(3 m+5)^{2}$
(ii) $(5 \mathrm{p}-1)^{2}$
(iii) $(2 \mathrm{n}-1)(2 \mathrm{n}+3)$
(iv) $4 p^{2}-25 q^{2}$
Answer:
(i) $(3 m+5)^{2}$
Comparing $(3 \mathrm{~m}+5)^{2}$ with $(\mathrm{a}+\mathrm{b})^{2}$ we have $\mathrm{a}=3 \mathrm{~m}$ and $\mathrm{b}=5$
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(3 m+5)^{2}=(3 m)^{2}+2(3 m)(5)+5^{2} \\
&=3^{2} m^{2}+30 m+25 \\
&=9 m^{2}+30 m+25
\end{aligned}$
(ii) $(5 p-1)^{2}$
Comparing $(5 p-1)^{2}$ with $(a-b)^{2}$ we have $a=5 p$ and $b=1$
$(a-b)^{2}=a^{2}-2 a b+b^{2}$ $(5 p-1)^{2}=(5 p)^{2}-2(5 p)(1)+1^{2}$ $=5^{2} p^{2}-10 p+1$ $=25 p^{2}-10 p+1$ $($ iii $)(2 n-1)(2 n+3)$ Comparing $(2 n-1)(2 n+3)$ with $(x+a)(x+b)$ we have $a=-1 ; b=3$ $(x+a)(x+b)=x^{2}+(a+b) x+a b$
Comparing $(2 n-1)(2 n+3)$ with $(x+a)(x+b)$ we have $a=-1 ; b=3$ $(x+a)(x+b)=x^{2}+(a+b) x+a b$

$\begin{aligned}
&(2 n+(-1))(2 n+3)=(2 n)^{2}+(-1+3) 2 n+(-1)(3) \\
&=2^{2} n^{2}+2(2 n)-3=4 n^{2}+4 n-3
\end{aligned}$
(iv) $4 p^{2}-25 q^{2}=(2 p)^{2}-(5 q)^{2}$
Comparing $(2 p)^{2}-(5 q)^{2}$ with $a^{2}-b^{2}$ we have $a=2 p$ and $b=5 q$
$\begin{aligned}
&\left(a^{2}-b^{2}\right)=(a+b)(a-b) \\
&=(2 p+5 q)(2 p-5 q)
\end{aligned}$

Question $2 .$
Expand
(i) $(3+\mathrm{m})^{3}$
(ii) $(2 a+5)^{3}$
(iii) $(3 p+4 q)^{3}$
(iv) $(52)^{3}$
(v) $(104)^{3}$
Answer:
(i) $(3+\mathrm{m})^{3}$
Cornparing $(3+m)^{3}$ with $(a+b)^{3}$ we have $a=3 ; b=m$
$\begin{aligned}
&(a+b)^{3}=a^{2}+3 a^{2} b+3 a b^{2}+b^{3} \\
&(3+m)^{3}=3^{3}+3(3)^{2}(m)+3(3) m^{2}+m^{3} \\
&=27+27 m+9 m^{2}+m^{3} \\
&=m^{3}+9 m^{2}+27 m+27
\end{aligned}$
(ii) $(2 \mathrm{a}+5)^{3}=$
Comparing $(2 \mathrm{a}+5)^{3}$ with $(\mathrm{a}+\mathrm{b})^{3}$ we have $\mathrm{a}=2 \mathrm{a}, \mathrm{b}=5$
$\begin{aligned}
&(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
&=(2 a)^{3}+3(2 a)^{2} 5+3(2 a) 5^{2}+5^{3} \\
&=2^{3} a^{3}+3\left(2^{2} a^{2}\right) 5+6 a(25)+125
\end{aligned}$

$=8 \mathrm{a}^{3}+60 \mathrm{a}^{2}+150 \mathrm{a}+125$
(iii) $(3 p+4 q)^{3}$
Comparing $(3 \mathrm{p}+4 \mathrm{q})^{3}$ with $(\mathrm{a}+\mathrm{b})^{3}$ we have $\mathrm{a}=3 \mathrm{p}$ and $\mathrm{b}=4 \mathrm{q}$
$\begin{aligned}
&(a+b)=a^{3}+3 a^{2} b+3 a b^{2}+b^{2} \\
&(3 p+4 q)^{3}=(3 p)^{3}+3(3 p)^{2}(4 q)+3(3 p)(4 q)^{2}+(4 q)^{3} \\
&=3^{3} p^{3}+3\left(9 p^{2}\right)(4 q)+9 p\left(16 q^{2}\right)-4^{3} q^{3} \\
&=27 p^{3}+108 p^{2} q+144 p q^{3}+64 q^{3}
\end{aligned}$
(iv) $(52)^{3}$
$(52)^{3}=(50+2)^{3}$
Comparing $(50+2)^{3}$ with $(a+b)^{3}$ we have $a=50$ and $b=2$
$\begin{aligned}
&(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
&(50+2)^{3}=50^{3}+3(50)^{2} 2+3(50)(2)^{2}+2^{3} \\
&52^{3}=125000+6(2,500)+150(4)+8 \\
&=1,25,000+15,000+600+8 \\
&52^{3}=1,40,608
\end{aligned}$

(v) $(104)^{3}$
$(104)^{3}=(100+4)^{3}$
Comparing $(100+4)^{3}$ with $(a+b)^{3}$ we have $a-100$ and $b=4$
$\begin{aligned}
&(\mathrm{a}+\mathrm{b})^{3}=\mathrm{a}^{3}+3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}{ }^{2}+\mathrm{b}^{3} \\
&(100+4)^{3}=(100)^{3}+3(100)^{2}(4)+3(100)(4)^{2}+4^{3} \\
&=10,00,000+3(10000) 4+300(16)+64 \\
&=10,00,000+1,20,000+4,800+64=11,24,864
\end{aligned}$

Question $3 .$
Expand
(i) $(5-x)^{3}$
(ii) $(2 x-4 y)^{3}$
(iii) $(a b-c)^{3}$
(iv) $(48)^{3}$
(v) $(97 x y)^{3}$
Answer:
(i) $(5-\mathrm{x})^{3}$
Comparing $(5-x)^{3}$ with $(a-b)^{3}$ we have $a=5$ and $b=x$
$(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$
$(5-x)^{3}=5^{3}-3(5)^{2}(x)+3(5)\left(x^{2}\right)-x^{3}$

$\begin{aligned}
&=125-3(25)(x)+15 x^{2}-x^{3} \\
&=125-75 x+15 x^{2}-x^{3}
\end{aligned}$
(ii) $(2 x-4 y)^{3}$
Comparing $(2 x-4 y)^{3}$ with $(a-b)^{3}$ we have $a=2 x$ and $b=4 y$
$\begin{aligned}
&(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\
&(2 x-4 y)^{3}=(2 x)^{3}-3(2 x)^{2}(4 y)+3(2 x)(4 y)^{2}-(4 y)^{3} \\
&=2^{3} x^{3}-3\left(2^{2} x^{2}\right)(4 y)+3(2 x)\left(4^{2} y^{2}\right)-\left(4^{3} y^{3}\right) \\
&=8 x^{3}-48 x^{2} y+96 x y^{2}-64 y^{3}
\end{aligned}$
(iii) $(a b-c)^{3}$
Comparing $(a b-c)^{3}$ with $(a-b)^{3}$ we have $a=a b$ and $b=c$
$\begin{aligned}
&(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\
&(a b-c)^{3}=(a b)^{3}-3(a b)^{2} c+3 a b(c)^{2}-c^{3} \\
&=a^{3} b^{3}-3\left(a^{2} b^{2}\right) c+3 a b c^{2}-c^{3} \\
&=a^{3} b^{3}-3 a^{2} b^{2} c+3 a b c^{2}-c^{3}
\end{aligned}$
(iv) $(48)^{3}$
$(48)^{3}=(50-2)^{3}$
Comparing $(50-2)^{3}$ with $(a-b)^{3}$ we have $a=50$ and $b=2$
$\begin{aligned}
&(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\
&(50-2)^{3}=(50)^{3}-3(50)^{2}(2)+3(50)(2)^{2}-2^{3} \\
&=1,25,000-15000+600-8 \\
&=1,10,000+592 \\
&=1,10,592
\end{aligned}$

(v) $(97 x y)^{3}$
$(97 x y)^{3}=97^{3} x^{3} y^{3}=(100-3) x^{3} y^{3} \ldots \text { (1) }$
Comparing $(100-3)^{3}$ with $(a-b)^{3}$ we have $a=100, b=3$
$(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$
$\begin{aligned}
&(100-3)^{3}=(100)^{3}-3(100)^{2}(3)+3(100)(3)^{2}-3^{3} \\
&97^{3}=10,00,000-90000+2700-27 \\
&97^{3}=910000+2673 \\
&97^{3}=912673 \\
&97 \mathrm{x}^{3} \mathrm{y}^{3}=912673 \mathrm{x}^{3} \mathrm{y}^{3}
\end{aligned}$

Question $4 .$
Simplify $(p-2)(p+1)(p-4)$
Answer:
$(\mathrm{p}-2)(\mathrm{p}+1)(\mathrm{p}-4)=(\mathrm{p}+(-2))(\mathrm{p}+1)(\mathrm{p}+(-4))$
Comparing $(\mathrm{p}-2)(\mathrm{p}+1)(\mathrm{p}-4)$ with $(\mathrm{x}+\mathrm{a})(\mathrm{x}+\mathrm{b})(\mathrm{x}+\mathrm{c})$ we have $\mathrm{x}=\mathrm{p} ; \mathrm{a}=-2$; $\mathrm{b}=1 ; \mathrm{c}=-4$.
$\begin{aligned}
&(x+a)(x+b)(x+c)=x^{2}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c \\
&\left.=p^{3}+(-2+1+(-4)) p^{2}+(-2)(1)+(1)(-4)+(-4)(-2)\right) p+(-2)(1)(-4) \\
&=p^{3}+(-5) p^{2}+(-2+(-4)+8) p+8 \\
&=p^{2}-5 p^{2}+2 p+8
\end{aligned}$

Question $5 .$
Find the volume of the cube whose side is $(x+1) \mathrm{cm}$
Answer:
Given side of the cube $=(x+1) \mathrm{cm}$
Volume of the cube $=(\text { side })^{3}$ cubic units $=(x+1)^{3} \mathrm{~cm}^{3}$
We have $(a+b)^{3}=\left(a 3^{3}+3 a^{2} b+3 a b^{2}+b^{3}\right){c m^{3}}^{3}$
$(x+1)^{3}=\left(x^{3}+3 x^{2}(1)+3 x(1)^{2}+1^{3}\right) \mathrm{cm}^{3}$
Volume $=\left(\mathrm{x}^{3}+3 \mathrm{x}^{2}+3 \mathrm{x}+1\right) \mathrm{cm}^{3}$
 

Question $6 .$
Find the volume of the cuboid whose dimensions are $(x+2),(x-1)$ and $(x-3)$
Answer:
Given the dimensions of the cuboid as $(x+2),(x-1)$ and $(x-3)$
$\therefore$ Volume of the cuboid $=(1 \times \mathrm{b} \times \mathrm{h})$ units $^{3}$
$=(x+2)(x-1)(x-3)$ units $^{3}$
We have $(x+a)(x+b)(x+c)=x^{3}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c$ $\therefore(\mathrm{x}+2)(\mathrm{x}-1)(\mathrm{x}-3)=\mathrm{x}^{3}+(2-1-3) \mathrm{x}^{2}+(2(-1)+(-1)(-3)+(-3)(2)) \mathrm{x}+(2)(-1)$
$(-3)$
$=x^{3}-2 x^{2}+(-2+3-6) x+6$
Volume $=x^{3}-2 x^{3}-5 x+6$ units $^{3}$

Objective Type Questions
Question 7 .
If $x^{2}-y^{2}=16$ and $(x+y)=8$ then $(x-y)$ is
(A) 8
(B) 3
(C) 2
(D) 1
Answer:
(C) 2
Hint:
$\begin{aligned}
&x^{2}-y^{2}=16 \\
&(x+y)(x-y)=16 \\
&8(x-y)=16 \\
&(x-y)=\frac{16}{8}=2
\end{aligned}$

Question 8 .
$\frac{(a+b)\left(a^{3}-b^{3}\right)}{\left(a^{2}-b^{2}\right)}=$
(A) $a^{2}-a b+b^{2}$
(B) $a^{2}+a b+b^{2}$
(C) $a^{2}+2 a b+b^{2}$
(D) $a 2^{2}-2 a b+b^{2}$
Answer:
(B) $a^{2}+a b+b^{2}$

Question 9 .
$(p+q)\left(p^{2}-p q+q^{2}\right)$ is equal to
(A) $\mathrm{p}^{3}+\mathrm{q}^{3}$
(B) $(p+q)^{3}$
(C) $\mathrm{p}^{3}-\mathrm{q}^{3}$
(D) $(p-q)^{3}$
Answer:
(A) $\mathrm{p}^{3}+\mathrm{q}^{3}$
Hint:
$a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$
 

Question $10 .$
$(a-b)=3$ and $a b=5$ then $a^{3}-b^{3}=$
(A) 15
(B) 18
(C) 62
(D) 72
Answer:
(D) 72
Hint:
$(a-b)=3$
$(a-b)^{2}=3^{2}$
$a^{2}+b^{2}-2 a b=9$
$a^{2}+b^{2}-2(5)=9$
$a^{2}+b^{2}=9+10$
$a^{2}+b^{2}=19$
$a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)=3(19+5)$
$=3(24)=72$

Question 11.
$a^{3}+b^{3}=(a+b)^{3}$

(A) $3 a(a+b)$
(B) $3 a b(a-b)$
(C) $-3 a b(a+b)$
(D) $3 a b(a+b)$
Answer:
(D) $3 a b(a+b)$
Hint:
$\begin{aligned}
&(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \\
&(a+b)^{3}-3 a^{2} b-3 a b^{3}=a^{3}+b^{3} \\
&(a+b)^{3}-3 a b(a+b)=a^{3}+b^{3}
\end{aligned}$