SELINA Solution Class 9 Chapter 7 Indices (Exponents) Exercise 7A

Question 1.1

Evaluate : 
${\displaystyle 3^3\times {\left(243\right)}^{-\frac{2}{3}}\times 9^{-\frac{1}{3}}}$

Sol:

${\displaystyle 3^3\times {\left(243\right)}^{-\frac{2}{3}}\times 9^{\frac{1}{3}}}$

= ${\displaystyle 3^3\times {(3\times 3\times 3\times 3\times 3)}^{-\frac{2}{3}}\times {(3\times 3)}^{-\frac{1}{3}}}$

= ${\displaystyle 3^3\times {\left(3^5\right)}^{-\frac{2}{3}}\times {\left(3^2\right)}^{\frac{1}{3}}}$

= ${\displaystyle 3^3\times 3^{-\frac{10}{3}}\times 3^{-\frac{2}{3}} ...[{\left(a^m\right)}^n=a^{mn}]}$

= ${\displaystyle 3^{3-\frac{10}{3}-\frac{2}{3}} [a^m\times a^n\times a^o=a^{m+n+o}]}$             
=${\displaystyle 3^{\frac{9-10-2}{3}}}$   

= ${\displaystyle 3^{\frac{9-12}{3}}}$

= ${\displaystyle 3^{-\frac{3}{3}}}$

= ${\displaystyle 3^{-1}}$
= ${\displaystyle \frac{1}{3}}$

Question 1.2

Evaluate :
${\displaystyle 5^{-4}\times {\left(125\right)}^{\frac{5}{3}}÷{\left(25\right)}^{-\frac{1}{2}}}$

Sol:

${\displaystyle 5^{-4}\times {\left(125\right)}^{\frac{5}{3}}÷{\left(25\right)}^{-\frac{1}{2}}}$

= ${\displaystyle 5^{-4}\times {(5\times 5\times 5)}^{\frac{5}{3}}÷{(5\times 5)}^{-\frac{1}{2}}}$

= ${\displaystyle 5^{-4}\times {\left(5^3\right)}^{\frac{5}{3}}÷{\left(5^2\right)}^{-\frac{1}{2}}}$

= ${\displaystyle 5^{-4}\times \left[{\left(5\right)}^{3\times \frac{5}{3}}\right]÷\left[{\left(5\right)}^{2\times -\frac{1}{2}}\right]}$

= ${\displaystyle \frac{5^{-4}\times 5^5}{5^{-1}}}$

= ${\displaystyle \frac{5^{5-4}}{5^{-1}}}$

= ${\displaystyle \frac{5^1}{5^{-1}}}$

= ${\displaystyle 5^{1-(-1)}}$

= ${\displaystyle 5^2}$
= 5 x 5
=25

Question 1.3

Evaluate :
${\displaystyle {\left(\frac{27}{125}\right)}^{\frac{2}{3}}\times {\left(\frac{9}{25}\right)}^{\frac{3}{2}}}$

Sol:

${\displaystyle {\left(\frac{27}{125}\right)}^{\frac{2}{3}}\times {\left(\frac{9}{25}\right)}^{\frac{3}{2}}}$

= ${\displaystyle {\left(\frac{3\times 3\times 3}{5\times 5\times 5}\right)}^{\frac{2}{3}}\times {\left(\frac{3\times 3}{5\times 5}\right)}^{-\frac{3}{2}}}$

= ${\displaystyle {\left[{\left(\frac{3}{5}\right)}^3\right]}^{\frac{2}{3}}\times {\left[{\left(\frac{3}{5}\right)}^2\right]}^{-\frac{3}{2}}}$

= ${\displaystyle {\left(\frac{3}{5}\right)}^{3\times \frac{2}{3}}\times {\left(\frac{3}{5}\right)}^{2\times -\frac{3}{2}}}$

= ${\displaystyle {\left(\frac{3}{5}\right)}^2\times {\left(\frac{3}{5}\right)}^{-3}}$

= ${\displaystyle {\left(\frac{3}{5}\right)}^{2-3}}$

= ${\displaystyle {\left(\frac{3}{5}\right)}^{-1}}$

= ${\displaystyle \frac{1}{\frac{3}{5}}=\frac{5}{3}}$ 

Question 1.4

Evaluate :
${\displaystyle 7^0\times {\left(25\right)}^{-\frac{3}{2}}-5^{-3}}$

Sol:

${\displaystyle 7^0\times {\left(25\right)}^{-\frac{3}{2}}-5^{-3}}$
= ${\displaystyle 7^0\times {(5\times 5)}^{-\frac{3}{2}}-5^{-3}}$

= ${\displaystyle 7^0\times {\left(5^2\right)}^{-\frac{3}{2}}-\frac{1}{5^3}}$

= ${\displaystyle 7^0\times \left[{\left(5\right)}^{2\times (-\frac{3}{2})}\right]-\frac{1}{5^3}}$

= ${\displaystyle 7^0\times 5^{-3}-\frac{1}{5^3}}$

= ${\displaystyle 1\times 5^{-3}-\frac{1}{5^3}}$

= ${\displaystyle \frac{1}{5^3}-\frac{1}{5^3}}$

= ${\displaystyle \frac{1-1}{5\times 5\times 5}}$

= ${\displaystyle \frac{0}{125}}$

= 0

Question 1.5

Evaluate : 
${\displaystyle {\left(\frac{16}{81}\right)}^{-\frac{3}{4}}\times {\left(\frac{49}{9}\right)}^{\frac{3}{2}}÷{\left(\frac{343}{216}\right)}^{\frac{2}{3}}}$

Sol:

${\displaystyle {\left(\frac{16}{81}\right)}^{-\frac{3}{4}}\times {\left(\frac{49}{9}\right)}^{\frac{3}{2}}÷{\left(\frac{343}{216}\right)}^{\frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}\right)}^{-\frac{3}{4}}\times {\left(\frac{7\times 7}{3\times 3}\right)}^{\frac{3}{2}}÷{\left(\frac{7\times 7\times 7}{6\times 6\times 6}\right)}^{\frac{2}{3}}}$

= ${\displaystyle {\left[{\left(\frac{2}{3}\right)}^4\right]}^{-\frac{3}{4}}\times {\left[{\left(\frac{7}{3}\right)}^2\right]}^{\frac{3}{2}}÷{\left[{\left(\frac{7}{6}\right)}^3\right]}^{\frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2}{3}\right)}^{4\times -\frac{3}{4}}\times {\left(\frac{7}{3}\right)}^{2\times \frac{3}{2}}÷{\left(\frac{7}{6}\right)}^{3\times \frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2}{3}\right)}^{-3}\times {\left(\frac{7}{3}\right)}^3÷{\left(\frac{7}{6}\right)}^2}$

= ${\displaystyle \frac{1}{{\left(\frac{2}{3}\right)}^3}\times {\left(\frac{7}{3}\right)}^3\times \frac{1}{{\left(\frac{7}{6}\right)}^2}}$

= ${\displaystyle \frac{1}{\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}}\times \frac{7}{3}\times \frac{7}{3}\times \frac{7}{3}\times \frac{1}{\frac{7}{6}\times \frac{7}{6}}}$

= ${\displaystyle \frac{1\times 3\times 3\times 3}{2\times 2\times 2}\times \frac{7}{3}\times \frac{7}{3}\times \frac{7}{3}\times \frac{1\times 6\times 6}{7\times 7}}$

= ${\displaystyle \frac{7\times 3\times 3}{2}}$

= ${\displaystyle \frac{63}{2}}$

= 31.5

Question 2.1

Simplify :
${\displaystyle {(8x^3÷125y^3)}^{\frac{2}{3}}}$

Sol:

${\displaystyle {(8x^3÷125y^3)}^{\frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2x\times 2x\times 2x}{5y\times 5y\times 5y}\right)}^{\frac{2}{3}}}$

= ${\displaystyle {\left[{\left(\frac{2x}{5y}\right)}^3\right]}^{\frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2x}{5y}\right)}^{3\times \frac{2}{3}}}$

= ${\displaystyle {\left(\frac{2x}{5y}\right)}^2}$

= ${\displaystyle [\frac{2x}{5y}\times \frac{2x}{5y}]}$

= ${\displaystyle \frac{4x^2}{25y^2}}$

Question 2.2

Simplify :
${\displaystyle {(a+b)}^{-1}.(a^{-1}+b^{-1})}$

Sol:

${\displaystyle {(a+b)}^{-1}.(a^{-1}+b^{-1})}$

= ${\displaystyle \frac{1}{a+b}\times (\frac{1}{a}+\frac{1}{b})}$

= ${\displaystyle \frac{1}{a+b}\times \left(\frac{b+a}{ab}\right)}$

= ${\displaystyle \frac{1}{a+b}\times \frac{a+b}{ab}}$

= ${\displaystyle \frac{1}{ab}}$

Question 2.3

Simplify :
${\displaystyle \frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^n-5^n\times 2^2}}$

Sol:

${\displaystyle \frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^n-5^n\times 2^2}}$

= ${\displaystyle \frac{5^{n+1}\times 5^2-6\times 5^{n+1}}{9\times 5^n-5^n\times 2^2}}$

= ${\displaystyle \frac{5^{n+1}\times (5^2-6)}{5^n\times (9-4)}}$

= ${\displaystyle \frac{5^1\times 19}{5}}$

= 19

Question 2.4

Simplify :
${\displaystyle {\left(3x^2\right)}^{-3}\times {\left(x^9\right)}^{\frac{2}{3}}}$

Sol:

${\displaystyle {\left(3x^2\right)}^{-3}\times {\left(x^9\right)}^{\frac{2}{3}}}$

= ${\displaystyle \frac{1}{{\left(3x^2\right)}^3}\times x^{9\times \frac{2}{3}}}$

= ${\displaystyle \frac{1}{3^3{x}^{2\times 3}}\times x^6}$

= ${\displaystyle \frac{1}{27x^6}\times x^6}$

= ${\displaystyle \frac{1}{27}}$

Question 3.1

Evaluate :
${\displaystyle \sqrt{\frac{1}{4}}+{\left(0.01\right)}^{-\frac{1}{2}}-{\left(27\right)}^{\frac{2}{3}}}$

Sol:

${\displaystyle \sqrt{\frac{1}{4}}+{\left(0.01\right)}^{-\frac{1}{2}}-{\left(27\right)}^{\frac{2}{3}}}$

= ${\displaystyle \sqrt{\frac{1}{2}\times \frac{1}{2}}+{(0.1\times 0.1)}^{-\frac{1}{2}}-{(3\times 3\times 3)}^{\frac{2}{3}}}$

= ${\displaystyle \frac{1}{2}+{\left[{\left(0.1\right)}^2\right]}^{-\frac{1}{2}}-{\left(3^2\right)}^{\frac{2}{3}}}$

= ${\displaystyle \frac{1}{2}+{\left(0.1\right)}^{2\times (-\frac{1}{2})}-3\times (3\times \frac{2}{3})}$

= ${\displaystyle \frac{1}{2}+{\left(0.1\right)}^{-1}-3^2}$

= ${\displaystyle \frac{1}{2}+\frac{1}{0.1}-9}$

= ${\displaystyle \frac{1}{2}+\frac{10}{1}-9}$

= ${\displaystyle \frac{1+20-18}{2}}$

= ${\displaystyle \frac{3}{2}}$

= ${\displaystyle 1\frac{1}{2}}$

Question 3.2

Evaluate :
${\displaystyle {\left(\frac{27}{8}\right)}^{\frac{2}{3}}-{\left(\frac{1}{4}\right)}^{-2}+5^0}$

Evaluate :
${\displaystyle {\left(\frac{27}{8}\right)}^{\frac{2}{3}}-{\left(\frac{1}{4}\right)}^{-2}+5^0}$

Sol:

${\displaystyle {\left(\frac{27}{8}\right)}^{\frac{2}{3}}-{\left(\frac{1}{4}\right)}^{-2}+5^0}$

= ${\displaystyle {\left(\frac{3\times 3\times 3}{2\times 2\times 2}\right)}^{\frac{2}{3}}-{\left(\frac{1\times 1}{2\times 2}\right)}^{-2}+5^0}$

= ${\displaystyle {\left[{\left(\frac{3}{2}\right)}^3\right]}^{\frac{2}{3}}-{\left[{\left(\frac{1}{2}\right)}^2\right]}^{-2}+1}$

= ${\displaystyle {\left(\frac{3}{2}\right)}^{3\times \frac{2}{3}}-{\left(\frac{1}{2}\right)}^{2\times (-2)}+1}$

= ${\displaystyle {\left(\frac{3}{2}\right)}^2-{\left(\frac{1}{2}\right)}^{-4}+1}$

= ${\displaystyle \frac{3}{2}\times \frac{3}{2}-\frac{1}{{\left(\frac{1}{2}\right)}^4}+1}$

= ${\displaystyle \frac{9}{4}-\frac{1}{\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}}+1}$

= ${\displaystyle \frac{9}{4}-\frac{1}{\frac{1}{16}}+1}$

= ${\displaystyle \frac{9}{4}-16+1}$

= ${\displaystyle \frac{9-64+4}{4}}$

= ${\displaystyle \frac{-51}{4}}$

Question 4.1

Simplify the following and express with positive index :
${\displaystyle {\left(\frac{3^{-4}}{2^{-8}}\right)}^{\frac{1}{4}}}$

Sol:

${\displaystyle {\left(\frac{3^{-4}}{2^{-8}}\right)}^{\frac{1}{4}}}$ 

= ${\displaystyle {\left(\frac{2^8}{3^4}\right)}^{\frac{1}{4}}}$

= ${\displaystyle \frac{{\left(2^8\right)}^{\frac{1}{4}}}{{\left(3^4\right)}^{\frac{1}{4}}}}$

= ${\displaystyle \frac{2^{8\times \frac{1}{4}}}{3^{4\times \frac{1}{4}}}}$

= ${\displaystyle \frac{2^2}{3}}$

= ${\displaystyle \frac{4}{3}}$

Question 4.2

Simplify the following and express with positive index :
${\displaystyle {\left(\frac{{27}^{-3}}{9^{-3}}\right)}^{\frac{1}{5}}}$

Sol:

${\displaystyle {\left(\frac{{27}^{-3}}{9^{-3}}\right)}^{\frac{1}{5}}}$

= ${\displaystyle {\left(\frac{9^3}{{27}^3}\right)}^{\frac{1}{5}}}$

= ${\displaystyle {\left[\frac{{\left(3^2\right)}^3}{{\left(3^3\right)}^3}\right]}^{\frac{1}{5}}}$

= ${\displaystyle {\left[{\left(\frac{3^2}{3^3}\right)}^3\right]}^{\frac{1}{5}}}$

= ${\displaystyle {\left[{\left(\frac{1}{3}\right)}^3\right]}^{\frac{1}{5}}}$

= ${\displaystyle {\left(\frac{1}{3}\right)}^{3\times \frac{1}{5}}}$

= ${\displaystyle \frac{1}{{\left(3\right)}^{\frac{3}{5}}}}$

Question 4.3

Simplify the following and express with positive index :
${\displaystyle {\left(32\right)}^{-\frac{2}{5}}÷{\left(125\right)}^{-\frac{2}{5}}}$

Sol:

${\displaystyle {\left(32\right)}^{-\frac{2}{5}}÷{\left(125\right)}^{-\frac{2}{5}}}$

= ${\displaystyle \left[\frac{{\left(32\right)}^{-\frac{2}{5}}}{{\left(125\right)}^{-\frac{2}{3}}}\right]}$

= ${\displaystyle \frac{{\left(125\right)}^{\frac{2}{3}}}{{\left(32\right)}^{\frac{2}{5}}}}$

= ${\displaystyle \frac{{(5\times 5\times 5)}^{\frac{2}{3}}}{{(2\times 2\times 2\times 2\times 2)}^{\frac{2}{5}}}}$

= ${\displaystyle \frac{{\left(5^3\right)}^{\frac{2}{3}}}{{\left(2^5\right)}^{\frac{2}{5}}}}$

= ${\displaystyle \frac{5^2}{2^2}}$

= ${\displaystyle \frac{25}{4}}$

= ${\displaystyle 6\frac{1}{4}}$

Question 4.4

Simplify the following and express with positive index :
${\displaystyle {[1-{\{1-{(1-n)}^{-1}\}}^{-1}]}^{-1}}$

Sol:

${\displaystyle {[1-{\{1-{(1-n)}^{-1}\}}^{-1}]}^{-1}}$

= ${\displaystyle \frac{1}{{[1-{\{1-{(1-n)}^{-1}\}}^{-1}]}^{+}1}}$

= ${\displaystyle \frac{1}{1-\frac{1}{1-{(1-n)}^{-1}}}}$

= ${\displaystyle \frac{1}{1- \frac{1}{1- \frac{1}{1-n}}}}$

= ${\displaystyle \frac{1}{1-\frac{\frac{1}{1(1-n)-1}}{1-n}}}$

= ${\displaystyle \frac{1}{1-\frac{\frac{1}{-n}}{1-n}}}$

= ${\displaystyle \frac{1}{1-\frac{1-n}{-n}}}$

= ${\displaystyle \frac{1}{1+\frac{1-n}{n}}}$

= ${\displaystyle \frac{1}{\frac{n+(1-n)}{n}}}$

= ${\displaystyle \frac{\frac{1}{n+1-n}}{n}}$

= ${\displaystyle \frac{n}{1}}$

= n

Question 5

If 2160 = 2^a. 3^b. 5^c, find a, b and c. Hence calculate the value of 3^a  x 2^-b x 5^-c.

Sol:

2160 = 2^a x 3^b x 5^c
⇒ 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 = 2^a x 3^b x 5^c
⇒ 2⁴ x 3³ x 5¹ = 2^a x 3^b x 5^c
⇒ 2^a x 3^b x 5^c = 2⁴ x 3³ x 5¹
Comparing powers of 2, 3 and 5 on the both sides of equation, We have
a = 4; b = 3 and c = 1
Hence,
Value of 3^a x 2^-b x 5^-c
= 3⁴ x 2^-3 x 5^-1
= 3 x 3 x 3 x 3 x ${\displaystyle \frac{1}{2^3}\times \frac{1}{5}}$ 

= ${\displaystyle 81\times \frac{1}{2\times 2\times 2}\times \frac{1}{5}}$

= ${\displaystyle 81\times \frac{1}{8}\times \frac{1}{5}}$

= ${\displaystyle \frac{81}{40}}$

= ${\displaystyle 2\frac{1}{40}}$

Question 6

If 1960 = 2^a. 5^b. 7^c, calculate the value of 2^-a. 7^b. 5^-c.

Sol:

1960 = 2^a x 5^b x 7^c
⇒ 2 x 2 x 2 x 5 x 7 x 7 = 2^a x 5^b x 7^c
⇒ 2³ x 5¹ x 7² = 2^a x 5^b x 7^c
⇒  2^ax 5^b x 7^c = 2³ x 5¹ x 7²
Comparing powers of 2,5 and 7 on the both sides of equation, We have
a = 3; b = 1 and c = 2
Hence,
Value of 2^-a x 7^b x 5^-c

= 2^-3 x 7¹ x 5^-2

= ${\displaystyle \frac{1}{2^3}\times 7\times \frac{1}{5^2}}$

= ${\displaystyle \frac{1}{8}\times 7\times \frac{1}{5\times 5}}$

= ${\displaystyle \frac{7}{200}}$          

Question 7.1

Simplify :
${\displaystyle \frac{8^{3}a\times 2^5\times 2^{2a}}{4\times 2^{11a}\times 2^{-2a}}}$

Sol:

${\displaystyle \frac{8^{3}a\times 2^5\times 2^{2a}}{4\times 2^{11a}\times 2^{-2a}}}$

= ${\displaystyle \frac{{\left(2^3\right)}^{3a}\times 2^5\times 2^{2a}}{2^2\times 2^{11a}\times 2^{-2a}}}$

=${\displaystyle \frac{2^{3\times 3a}\times 2^5\times 2^{2a}}{2^2\times 2^{11a}\times 2^{-2a}}}$

= ${\displaystyle \frac{2^{9a}\times 2^5\times 2^{2a}}{2^2\times 2^{11a}\times 2^{-2a}}}$

= ${\displaystyle 2^{9a+5+2a-2-11a+2a}}$

= ${\displaystyle 2^{2a+3}}$

Question 7.2

Simplify :
${\displaystyle \frac{3\times {27}^{n+1}+9\times 3^{3n-1}}{8\times 3^{3n}-5\times {27}^n}}$

Sol:

${\displaystyle \frac{3\times {27}^{n+1}+9\times 3^{3n-1}}{8\times 3^{3n}-5\times {27}^n}}$

= ${\displaystyle \frac{3\times {(3\times 3\times 3)}^{n+1}+3\times 3\times 3^{3n-1}}{2\times 2\times 2\times 3^{3n}-5\times {(3\times 3\times 3)}^n}}$

= ${\displaystyle \frac{3\times {\left(3^3\right)}^{n+1}+3^2\times 3^{3n-1}}{2^3\times 3^{3n}-5\times {\left(3^3\right)}^n}}$

= ${\displaystyle \frac{3\times 3^{3n+3}+3^{3n+1}}{2^3\times {\left(3^3\right)}^n-5\times {\left(3^3\right)}^n}}$

= ${\displaystyle \frac{3^{3n+3+1}+3^{3n+1}}{2^3\times {\left(3^3\right)}^n-5\times {\left(3^3\right)}^n}}$

= ${\displaystyle \frac{3^{3n+4}+3^{3n+1}}{2^3\times {\left(3^3\right)}^n-5\times {\left(3^3\right)}^n}}$

= ${\displaystyle \frac{3^{3n}\times 3^4+3^{3n}\times 3^1}{2^3\times {\left(3^3\right)}^n-5\times {\left(3^3\right)}^n}}$

= ${\displaystyle \frac{3^{3n}(3^4+3^1)}{{\left(3^3\right)}^n(8-5)}}$

= ${\displaystyle \frac{3^{3n}(3^4+3^1)}{3^{3n}\times 3}}$

= ${\displaystyle \frac{3\times 3\times 3\times 3+3}{3}}$

= ${\displaystyle \frac{81+3}{3}}$

= ${\displaystyle \frac{84}{3}}$

= 28

Question 8

Show that :
${\displaystyle {\left(\frac{a^m}{a^{-n}}\right)}^{m-n}\times {\left(\frac{a^n}{a^{-l}}\right)}^{n-l}\times {\left(\frac{a^l}{a^{-m}}\right)}^{l-m}=1}$

Sol:

${\displaystyle {\left(\frac{a^m}{a^{-n}}\right)}^{m-n}\times {\left(\frac{a^n}{a^{-l}}\right)}^{n-l}\times {\left(\frac{a^l}{a^{-m}}\right)}^{l-m}=1}$

= ${\displaystyle {(a^m\times a^n)}^{m-n}\times {(a^n\times a^l)}^{n-l}\times {(a^l\times a^m)}^{l-m}}$

= ${\displaystyle {\left(a^{m+n}\right)}^{m-n}\times {\left(a^{n+l}\right)}^{n-l}\times {\left(a^{l+m}\right)}^{l-m}}$

= ${\displaystyle a^{m^2-n^2}\times a^{n^2-l^2}\times a^{l^2-m^2}}$

= ${\displaystyle a^{m^2-n^2+n^2-l^2+l^2-m^2}}$

= ${\displaystyle a^0}$
= 1

Question 9

If a = x^{m + n}. y^l ; b = x^n + l. y^m and c = x^{l + m}. yⁿ,

Prove that : a^{m - n}. b^n - l. c^{l - m} = 1

Sol:

a = x^{m + n}. y^l
b = x^n + l. y^m
c = x^{l + m}. yⁿ

a^{m - n}. b^n - l. c^{l - m} = 1
LHS

= a^{m - n}. b^n - l. c^{l - m}

 ⁼  [ x^{( m + n ).} y^l ]^{( m - n )} . [ x^{( n + l )}. y^m ]^{( n - l )} . [ x^{( l + m )} . yⁿ ]^{( l - m )}

= x^{( m + n )( m - n )}. y^{l( m - n )} . x^{( n + l )( n - l )}. y^{m( n - l )} . x^{( l + m )( l - m )}. y^{n( l - m )}

= ${\displaystyle x^{m^2-n^2+n^2-l^2+l^2-m^2}.y^{lm-\ln +mn-ml+nl-nm}}$

= ${\displaystyle x^0.y^0}$

= 1 
= RHS

Question 10.1

Simplify : 
${\displaystyle {\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times {\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times {\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2}}$

Sol:

${\displaystyle {\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times {\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times {\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2}}$

= ${\displaystyle {\left(x^{a-b}\right)}^{a^2+ab+b^2}\times {\left(x^{b-c}\right)}^{b^2+bc+c^2}\times {\left(x^{c-a}\right)}^{c^2+ca+a^2}}$

= ${\displaystyle x^{a^3-b^3}\times x^{b^3-c^3}\times x^{c^3-a^3}}$

= ${\displaystyle x^{a^3-b^3+b^3-c^3+c^3-a^3}}$

= ${\displaystyle x^0}$

= 1

Question 10.2

Simplify :
${\displaystyle {\left(\frac{x^a}{x^{-b}}\right)}^{a^2-ab+b^2}\times {\left(\frac{x^b}{x^{-c}}\right)}^{b^2-bc+c^2}\times {\left(\frac{x^c}{x^{-a}}\right)}^{c^2-ca+a^2}}$

Sol:

${\displaystyle {\left(\frac{x^a}{x^{-b}}\right)}^{a^2-ab+b^2}\times {\left(\frac{x^b}{x^{-c}}\right)}^{b^2-bc+c^2}\times {\left(\frac{x^c}{x^{-a}}\right)}^{c^2-ca+a^2}}$

= ${\displaystyle {\left(x^{a+b}\right)}^{a^2-ab+b^2}\times {\left(x^{b+c}\right)}^{b^2-bc+c^2}\times {\left(x^{c+a}\right)}^{c^2-ca+a^2}}$

= ${\displaystyle x^{a^3+b^3}\times x^{b^3+c^3}\times \left(x^{c^3+a^3}\right)}$

= ${\displaystyle x^{a^3+b^3+b^3+c^3+c^3+a^3}}$

= ${\displaystyle x^{(a^3+b^3+b^3+c^3+c^3+a^3)}}$

= ${\displaystyle x^{2(a^3+b^3+c^3)}}$