Exercise 6 C
Question 1
(i) $5^{0} \times 4^{-1}+8^{\tfrac{1}{3}}$
(ii) $\sqrt[3]{(64)^{-4}(125)^{-2}}$
(iii) $\left(9^{-3} \times 16^{3 / 2}\right)^{1 / 6}$
(iv) $\sqrt[3]{(16)^{-3 / 4} \times(125)^{-2}}$
(v) $(32)^{-2 / 5} \div(216)^{-2 / 3}$
Sol :
(i) $5^{0} \times 4^{-1}+8^{\tfrac{1}{3}}$
$=1 \times \frac{1}{4}+(2 \times 2 \times 2)^{\tfrac{1}{3}}$
$=\frac{1}{4}+\left(2^{3}\right)^{1 / 3}=\frac{1}{4}+\left(2^{3 \times \tfrac{1}{3}}\right)$
$=\frac{1}{4}+2^{1}$
$=2+\frac{1}{4}$
$\Rightarrow 2 \frac{1}{4}$
(ii) $\sqrt[3]{(64)^{-4}(125)^{-2}}$
$=\left[(64)^{-4}(125)^{-2}\right]^{1 / 3}$
$=\left[(64)^{-4 \times 1 / 3}\right]_{\times}\left[(125)^{-2 \times 1 / 3}\right]$
$=\left[(64)^{-4 / 3}\right] \times(125)^{-2 / 3}$
$=\left(4^{3}\right)^{-4 / 3} \times\left(5^{3}\right)^{-2 / 3}$
$=4^{3 \times(-4 / 3)} \times 5^{3 \times(-2 / 3)}$
$=4^{-4} \times 5^{-2}$
$=\frac{1}{4^{4} \times 5^{2}}=\frac{1}{4 \times 4 \times 4 \times 4 \times 5 \times 5}$
$=\frac{1}{6400}$
(iii) $\left(9^{-3} \times 16^{3 / 2}\right)^{1 / 6}$
=$\left[\left\{13)^{2}\right]^{-3} \times(24)^{3 / 2}\right]^{1 / 6}$
=$\left[(3)^{-6} \times(2)^{6}\right]^{1 / 6}$
=$3^{-6 \times 1 / 6} \times 2^{6} \times 1 / 6$
=$3^{-1} \times 2^{1}=\frac{1}{3} \times 2$
$=\frac{2}{3}$
(iv) $\sqrt[3]{(16)^{-3 / 4} \times(125)^{-2}}$
=$\sqrt[3]{\left(2^{4}\right)^{-3 / 4} \times\left(5^{3}\right)^{-2}}$
$=\left[2^{4 \times(-3 / 4)} \times 5^{3 \times(-2)}\right]^{1 / 3}$
=$\left[2^{-3} \times 5^{-6}\right]^{1 / 3}$
$\left[\frac{1}{2^{3} \times 5^{6}}\right]^{1 / 3}$
$\frac{1}{\left(2^{3}\right)^{1 / 2} \times\left(5^{6}\right)^{1 / 3}}=\frac{1}{2^{1} \times 5^{2}}$
$=\frac{1}{2 \times 5 \times 5}=\frac{1}{50}$
(v) $(32)^{-2 / 5} \div(216)^{-2 / 3}$
$=\left(2^{5}\right)^{-2 / 5} \div\left(6^{3}\right)^{-2 / 3}$
$=(2)^{-2 / 5} \times 5+(6)^{3 \times(-2 / 3)}$
$=2^{-2} \div 6^{-2}$
$=\left(\frac{2}{6}\right)^{-2}=\left(\frac{6}{2}\right)^{2}=\frac{6 \times 6}{2 \times 2}$
= $\frac{36}{4}$
=9
Question 2
(i) $\left(\frac{4}{125}\right)^{-2 / 3}$
(ii) $9^{3 / 2}-3 \times(5)^{0}-\left(\frac{1}{81}\right)^{-1 / 2}$
(iii) $\left(\frac{1}{4}\right)^{-2}-3$ $(8)^{2} / 3$ х $4^{0}+\left(\frac{3}{16}\right)^{-1 / 2}$
(iv) $16^{3 / 4}+2\left(\frac{1}{2}\right)^{-1} \times 3^0$
(v) $(81)^{3 / 4}-\left(\frac{1}{32}\right)^{-2 / 5}+(8)^{1 / 3}\left(\frac{1}{2}\right)^{-2}(2)^{0}$
Sol:
(i) $\left(\frac{4}{125}\right)^{-2 / 3}$
$\frac{1}{\left(\frac{265}{625}\right)^{1 / 4}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}$
=$\left(\frac{4}{5^{3}}\right)^{-2 / 3} \div \frac{1}{\left(\frac{4^{4}}{5^{4}}\right)^{1 / 2}}$ $+\left[\frac{\left(5^{2}\right)^{1 / 2}}{\left(4^{3}\right)^{1 / 3}}\right]^{0}$
$\left(\frac{4}{5}\right)^{3 \times(-2 / 3)} \times\left(\frac{4}{5}\right)^{4 \times 1 / 4}+1$
$=\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{1}+1$
$=\left(\frac{5}{4}\right)^{2} \times\left(\frac{5}{4}\right)^{-1}+1$
$=\left(\frac{5}{4}\right)^{2-1}+1$
$=\left(\frac{5}{4}\right)+1$
$=\frac{5}{4}+1=1 \frac{1}{4}+1$
$=2 \frac{1}{4}$ Ans
(ii) $9^{3 / 2}-3 \times(5)^{0}-\left(\frac{1}{81}\right)^{-1 / 2}$
$=\left(3^{2}\right)^{3 / 2}-3 \times 1-\left(\frac{1}{3^{4}}\right)^{-1 / 2}$
$=3^{2} \times 3 / 2-3-\frac{1}{\left(3^{4}\right)^{-1 / 2}}$
$=3^{3}-3-\frac{1}{(3)^{-2}}$
$=27-3-(3)^{2}=27-3-9$
=27-12
=15 Ans
(iii) $\left(\frac{1}{4}\right)^{-2}-3$ $(8)^{2} / 3$ х $4^{0}+\left(\frac{3}{16}\right)^{-1 / 2}$
=$\left(\frac{1}{2^{2}}\right)^{-2}-3\left(2^{3}\right)^{2 / 3} \times 1+\left[\frac{(3)^{2}}{(4)^{2}}\right]^{-1 / 2}$
$=\frac{1}{(2)^{2(-2)}}$ $-3\left(2^{3 \times 2 / 3}\right)$× 1+ $\left(\frac{3}{4}\right)^{2}$^$(-1 / 2)$
= $\frac{1}{2^{-4}}$ $-3 \times 2^{2} \times 1+\left(\frac{3}{4}\right)^{-1}$
$=2^{4}-3 \times 4+\frac{4}{3}=16-12+\frac{4}{3}$
$=4+1 \frac{1}{3}=5 \frac{1}{3}$
(iv) $16^{3 / 4}+2\left(\frac{1}{2}\right)^{-1} \times 3$
$=\left(2^{4}\right)^{3 / 4}+2 \times 2^{1} \times 1=2^{3}+2 \times 2 \times 1$
$=2 \times 2 \times 2+2 \times 2=8+4=12$ Ans
(v) $(81)^{3 / 4}-\left(\frac{1}{32}\right)^{-2 / 5}+(8)^{1 / 3}\left(\frac{1}{2}\right)^{-2}(2)^{0}$
$=\left(3^{4}\right)^{3 / 4}-\frac{1}{2^{5(-2 / 5)}}+\left(2^{3 \times 1 / 3}\right) \times \frac{1}{(2)^{-2}} \times 1$
$=(3)^{3}-\frac{1}{(2)^{-2}}+2^{2} \times \frac{2^{2}}{1}$
$=27-(2)^{1}+2 \times 2^{2}$
=27-4+8
=35-4
=31 Ans
Question 3
(a) Evaluate : $x^{1 / 2} \times y^{-1} \times z^{2 / 3}$ , when x = 9, y = 2, and z = 8.
(b) Evaluate as a fraction :
$\left(\frac{27}{8}\right)^{2 / 3}-\left(\frac{1}{4}\right)^{-2}+(5)^{0}$
(c) Evaluate as a fraction :
$\sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}-(27)^{2 / 3}$
Sol:
(a) $x^{1 / 2} \times y^{-1} \times z^{2 / 3}$
x=9
y=2
z=8
$\Rightarrow(9)^{1 / 2} \times(2)^{-1} \times(8)^{2 / 3}$
$=\left(3^{2}\right)^{1 / 2} \times \frac{1}{2} \times\left(2^{3}\right)^{2 / 3}$
$=3^{2} \times 1 / 2 \times \frac{1}{2} \times 2^{3 \times 2 / 3}$
$=3^{1} \times \frac{1}{2} \times 2^{2}=3 \times \frac{1}{2} \times 4$
$=3 \times 2=6$ Ans
(b) $\left(\frac{27}{8}\right)^{2 / 3}-\left(\frac{1}{4}\right)^{-2}+(5)^{0}$
$=\left(\frac{3^{3}}{2^{3}}\right)^{2 / 3}-\left(\frac{1}{2^{2}}\right)^{-2}+1$
$\frac{3^{3 \times 2 / 3}}{2^{3 \times 2 / 3}}-\frac{1}{2^{1 \times(-2)}}+1$
$=\frac{3^{2}}{2^{2}}-\frac{1}{2^{-4}}+1=\frac{9}{4}-2^{4}+1$
$=\frac{9}{4}-16+1=2 \frac{1}{4}-16+1$
=$-12 \frac{3}{4}$ Ans.
(c) $\sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}-(27)^{2 / 3}$
= $\left(\frac{1}{2^{2}}\right)^{1 / 2}+(0.1)^{2 \times(-1 / 2)}-\left(3^{3}\right)^{2 / 3}$
$=\frac{1}{2^{2 \times 1 / 2}}+(0.1)^{-1}-3^{2}$
$=\frac{1}{2}+\frac{1}{(0.1)}-9=\frac{1}{2}+\frac{10}{1}-9$
$=10 \frac{1}{2}-9=1 \frac{1}{2}$ Ans
Question 4
(i) $\frac{(64)^{5 n / 6}(27)^{-n / 6}}{(12)^{-n / 2}} \cdot 2^{6 n}$
(ii) $\frac{6 \cdot(8)^{n+1}+162^{3 n \cdot 2}}{10 \cdot 2^{3 n+1}-7 \cdot(8)^{n}}$
(iii) $\frac{4^{2 n} \cdot 2^{n+1}}{2^{n-3} \cdot 4^{2 n+1}}$
(iv) $\frac{(5)^{2 n+3}-(25)^{n+2}}{\left((125)^{n+1}\right]^{2 / 3}}$
Sol:
(i) $\frac{(64)^{5 n / 6}(27)^{-n / 6}}{(12)^{-n / 2}} \cdot 2^{6 n}$
=$\frac{\left(2^{6}\right)^{5 n / 6}\left(3^{9}\right)^{-n / 6}}{\left(2^{2} \times 3\right)^{-n / 2}} \cdot 2^{6 n}$
= $\frac{(2)^{6 \times \frac{5 n}{6}} \cdot(3)^{3 \times\left(\frac{-n}{6}\right)}}{(2)^{-\frac{n}{2} \times 2} \cdot(3)^{-\frac{n}{2}}} \cdot 2^{6 n}$
=$\frac{2^{5 n} \cdot 3^{-n / 2}}{2^{-n} \cdot 3^{-n / 2}} \cdot 2^{6 n}=(2)^{5 n+6 n+n} \cdot(3)^{-\frac{n}{2}}+\frac{n}{2}$
=$=2^{12 n} \cdot 3^{0}=(12)^{12 n} \cdot 1 .$
$=(2)^{12 n}$
(ii) $\frac{6 \cdot(8)^{n+1}+162^{3 n \cdot 2}}{10 \cdot 2^{3 n+1}-7 \cdot(8)^{n}}$
=$\frac{6 \cdot\left(2^{3}\right)^{n+1}+16 \cdot 2^{3 n-2}}{10 \cdot 2^{3 n+1}-7 \cdot\left(2^{3}\right)^{n}}$
= $\frac{6 \cdot 2^{3 n+3}+16 \cdot 2^{3 n-2}}{10 \cdot 2^{3 n+1}-7 \cdot 2^{3 n}}$
=$\frac{6 \cdot 2^{3 n} \cdot 2^{3}+16 \cdot 2^{3 n} \cdot 2^{-2}}{10 \cdot 2^{3 n} \cdot 2-(7 \cdot 2)^{3 n}}$
=$\frac{2^{3 n}\left(6 \times 2^{3}+16 \times 2^{-2}\right)}{2^{3 n}(10 \times 2-7)}$
=$\frac{6 \times 8+16 \times \frac{1}{4}}{20-7}$
$=\frac{48+4}{13}=\frac{52}{13}=4$
(iii) $\frac{4^{2 n} \cdot 2^{n+1}}{2^{n-3} \cdot 4^{2 n+1}}$
$=\frac{\left(2^{2}\right)^{2 n} \cdot 2^{n+1}}{2^{(n-3)}\left(2^{2}\right)^{2 n-1}}$
=$=\frac{2^{4 n} \cdot 2^{n+1}}{2^{n-3} \cdot 2^{4 n+2}}=\frac{2^{4 n+n+1}}{2^{n-3+4 n+2}}$
=$\frac{25 n+1}{25 n-1}$
$=25 n+1-5 n+1$
$=2^{2}$
=4
(iv) $\frac{(5)^{2 n+3}-(25)^{n+2}}{\left((125)^{n+1}\right]^{2 / 3}}$
=$\frac{5^{3 n+3}-(5)^{m+2}}{\left[\left(5^{3}\right)^{n+3}\right]^{2 / 3}}$
=$\frac{5^{2 n} \cdot 5^{3} \cdot 5^{2 n} \cdot 5^{4}}{5(3 n+5)^{3 / 2}}$
$\Rightarrow \frac{5^{2 n} \cdot 5^{3}-5^{2 n} \cdot 5^{4}}{5^{2 n} \cdot 5^{2}}$
=$\frac{5^{2 n}\left(5^{3}-5^{4}\right)}{5^{2} n \cdot 5^{2}}$
=$\frac{125-625}{25}$
$=\frac{-500}{25}$
=-20 Ans
Question 5
If $\frac{49^{n+1} \cdot 7^n -(343)^n}{7^{3m} \cdot 2^n}=\frac{3}{343}$ , prove that m = (m + 1)
Sol:
$\begin{aligned} & \frac{49^{n+1} \cdot 7^{n}-(343)^{n}}{7^{3 m} \cdot 2^{n}} \Rightarrow \frac{3}{343} \\ \Rightarrow & \frac{\left(7^{2}\right)^{n+1} \cdot 7^{n}-\left(7^{3}\right)^{n}}{7^{3 m} \cdot 2^{n}}=\frac{3}{343} \\ \Rightarrow & \frac{7^{n}+2 \cdot 7^{n}-7^{3 n}}{7^{3 m} \cdot 2^{4}}-\frac{3}{343} \\ \Rightarrow & \frac{7^{3 n}\left(7^{2}-1\right)}{16 \cdot 7^{3 m}}=\frac{3}{343} \\ \Rightarrow & \frac{7^{3 n} \times 48}{16 \cdot 7^{3 m}}=\frac{3}{343} \end{aligned}$
=$\frac{7^{3 n}}{7^{3 m}}=\frac{3 \times 16}{343 \times 48}=\frac{1}{343}$
=$\frac{\left(7^{n}\right)^{3}}{\left(7^{m}\right)^{3}}=\frac{1}{7^{3}}=\left(\frac{1}{7}\right)^{3} \quad \Rightarrow \frac{7^{n}}{7^{m}}=\frac{1}{7} 7.7 n=7^{m}$
$\Rightarrow \quad 7^{n+1}=7^{m}$ (comparing both sides) $\Rightarrow m=n+1$
Hence proved.
Question 6
If $2205=3^{a} \times 5^{b} \times 7^{c}$
(i) Find the numerical values of a, b and c
(ii) Hence evaluate $3^{a} \times 5^{-b} \times 7^{-c}$
Sol:
(i) $2205=3^{a} \times 5^{b} \times 7^{c}$
$\begin{array}{l|l}3 & 2205 \\\hline 3 & 735 \\\hline 5 & 245 \\\hline 7 & 49 \\\hline 7 & 7 \\\hline 1\end{array}$
On comparing = a=2
b=1
c=2
so $3^{2} \times 5^{1} \times 7^{2}=3^{a} \times 5^{b} \times 7^{c}$
(ii)
$\begin{aligned} & 3^{a} \times 5^{-b} \times 7^{-c} \\=& \frac{3^{a}}{5^{b} \times 7^{c}}=\frac{3^{2}}{5^{1} \times 7^{2}} \\=& \frac{9}{5 \times 49} \Rightarrow \frac{9}{245} \\ & \frac{9}{245} \text { Ans. } \end{aligned}$
Question 7
If $\left[\frac{b^3c^{-2}}{b^{-4}c^3}\right]^{-3}+\left[\frac{b^{-1}c}{b^2.c^{-2}}\right]^5=b^x.c^x$ , prove that x + y + 6 = 0
Sol:
$\left[\frac{b^{3} c^{-2}}{b^{-4} c^{3}}\right]^{-3} \div\left[\frac{b^{-1} c}{b^{2} \cdot c^{-2}}\right]^{5}=b^{x} \cdot c^{y}$
Now $\text { R.H.S. }=x+y+6=-6+0+6=0$
Hence proved.
Question 8
Prove that
(i) $(x+y)^{-1}\left(x^{-1}+y-1\right)=\frac{1}{x y}$
(ii) $\left(x^{-1}+y-1\right)^{-1}=\frac{x y}{x+y}$
(iii) $\frac{a+b+c}{a^{-1} b^{-1}+b^{-2} c^{-1}+c^{-1} a^{-1}}=a b c$
(iv) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$
(v) $a b \sqrt{\frac{x^{a}}{x^{b}}} \cdot b c \sqrt{\frac{x^{b}}{x^{c}}} \cdot \sqrt[ca]{\frac{x^{2}}{x^{a}}}=1$
Sol:
(i)
$\begin{aligned}(x&+y)^{-1}\left(x^{-1}+y-1\right)=\frac{1}{x y} \\ \text { L.H.S. } &=(x+y)^{-1}\left(x^{-1}+y-2\right) \\ &=\frac{1}{x+y} \times\left(\frac{1}{x}+\frac{1}{y}\right) \\ &=\frac{1}{x+y} \times \frac{y+5}{x y} \end{aligned}$
$=\frac{1}{x+y} \times \frac{x+y}{x y} \Rightarrow \frac{1}{x y}$
Hence pruerd,
$\text { L.H.S }=\text { R.H.S. }$
(ii) $\left(x^{-1}+y-1\right)^{-1}=\frac{x y}{x+y}$
L.H.S
$\begin{aligned} &\left(x^{-1}+y^{-1}\right)^{-1} \\=&\left[\frac{1}{x}+\frac{1}{y}\right]^{-1} \\=&\left[\frac{y+x}{x y}\right]^{-1}=\frac{x y}{x+y} \quad \text { R.H.S. } \end{aligned}$
(iii) $\frac{a+b+c}{a^{-1} b^{-1}+b^{-2} c^{-1}+c^{-1} a^{-1}}=a b c$
L.H.S
$\frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}=\frac{a+b+c}{\frac{1}{a} \frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c} \frac{1}{a}}$
$=\frac{a+b+c}{\frac{c+a+b}{a b c}} \Rightarrow(a+b+c) \times \frac{a b c}{(a+b+c)}$
=$a b c$ R.H.S (Hence proved)
(iv) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$
L.H.S
$=\frac{1}{x^{a-a}+x^{b-a} \cdot x^{c-a}}+\frac{1}{x^{b-b}+x^{a-b}+x^{c-b}}+\frac{1}{x^{c-c}+x^{b-c}+x^{a-c}}$
$=\frac{1}{x^{-a}\left(x^{a}+x^{b}+x^{c}\right)}+\frac{1}{x^{-b}\left(x^{b}+x^{a}+x^{c}\right)}+\frac{1}{x^{-c}\left(x^{2}+x^{a}+x^{b}\right)}$
$=\frac{1}{\left(x^{a}+x^{b}+x^{c}\right)}\left[\frac{1}{x^{-a}}+\frac{1}{x^{-b}}+\frac{1}{x-c}\right]$
$\frac{1}{\left(x^{a}+-x^{b}+x^{2}\right)}\left[\frac{x^{a}+x^{b}+x^{c}}{1}\right]=1$
L.H.S =R.H.S PROVED
(v) $a b \sqrt{\frac{x^{a}}{x^{b}}} \cdot b c \sqrt{\frac{x^{b}}{x^{c}}} \cdot \sqrt[ca]{\frac{x^{2}}{x^{a}}}=1$
$x^{\frac{a-b}{a b}} \cdot x \frac{b-c}{b c} \cdot x^{\frac{c-a}{c a}}$
$=x \frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}$
$x \frac{a b c-b c + a b-a c+b c-a b}{a b c}$
$x \frac{D}{a b c}$
$\begin{aligned}&=x^{0}=1 \\&L \cdot H \cdot S=R \cdot H \cdot S\end{aligned}$
Hence, proved
Question 9
(i) $\left(p^{1 / 3}-p^{-1/ 3}\right)\left(p^{2 / 3}+1+p^{-2 / 3}\right)$
(ii) $(\sqrt{11}+\sqrt{3})^{1 / 3}(\sqrt{11}-\sqrt{3})^{1 / 3}$
Sol:
(i)$\left(p^{1 / 3}-p^{-1/ 3}\right)\left(p^{2 / 3}+1+p^{-2 / 3}\right)$
if $p^{1 / 3}=x$ and $p^{-1 / 3}=y$
$=\left(a^{2}-y^{2}\right)\left(x^{2}+x y+y^{2}\right)$
$=x^{3}-y^{3}=\left(p^{1 / 3}\right)^{3}-\left(p^{-1 / 3}\right)^{3}$ $=p^{1}-p^{-1}=p-\frac{1}{p}$ Ans.
(ii) $(\sqrt{11}+\sqrt{3})^{1 / 3}(\sqrt{11}-\sqrt{3})^{1 / 3}$
if $\sqrt{11}=x$
$\sqrt{3}=y$
$=(x+y)^{1 / 3}(x-y)^{1 / 3}$
$[$ from formula. $]$= $\left(x^{2}-y^{2}\right)^{1 / 3}$
$=\left[(\sqrt{11})^{2}-(\sqrt{3})^{2}\right]^{1 / 3}$
$=(11-3)^{1 / 3}=(8)^{1 / 3}$
$=\left(2^{3}\right)^{1 / 3}=2^{3 / \frac{1}{3}}$
$=2^{2}=2$ Ans
Question 10
If $3^{x}=5^{y}=45^{z}$ , prove that $x =\frac{2yz}{y-z}$
Sol:
L.H.S
Suppose $3^{x}=5^{y}=45^{z}=a$
so, $\begin{aligned} a^{1 / x} &=3, a^{1 / y}=5, \frac{1}{a^{2}}=45 \\ & 45=3 \times 3 \times 5 \\ & \Rightarrow 3^{2} \times 5=45 \\ &=a^{\frac{1}{z}}=a^{\frac{2}{x}} \times a^{1 / y} \\ &=a^{\frac{1}{z}}=a^{\frac{2}{x}+\frac{1}{y}} \end{aligned}$
In comparing
$\quad \frac{1}{2}=\frac{2}{x}+\frac{1}{y}=\frac{2}{x}=\frac{1}{z}-\frac{1}{y}$ $\Rightarrow \quad \frac{2}{x}=\frac{y-z}{y z}$ $\Rightarrow \frac{1}{x}=\frac{y-z}{2 y z}$ $\Rightarrow \quad x=\frac{2 y z}{y-z}$ R.H.S.
Hence proved
Question 11
If $x = ab^{p-1}, y = ab^{q-1}$ and $z = ab^{r-1}$ prove that $x^{p-r}x y^{r-p}.z^{p-q} = 1$.
Sol;
L.H.S.
$\begin{aligned}&x=a b^{p-1} \\&y=a b^{q-1} \\&z=a b^{r-1}\end{aligned}$
So $=x^{-r-r} \cdot y^{r-p} \cdot z^{p-q}$
$=\left(a b^{p-1}\right)^{q-\dot{r}} \cdot\left(a b^{2-1}\right)^{r-p} \cdot\left(a b^{r-1}\right)^{p-q}$
$=a^{q-r} \cdot b^{(2-r)(p-1)} \cdot a^{r-p} \cdot b^{(r-p)(r-1)} \cdot a^{p-2} \cdot b^{(p-2)}(r-1)$
$=a^{\circ} \cdot b^{\circ}$
$=1 \times 1=1$ R.H.S
Hence proved
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