Question

If ( a - 1/a ) = 8/3 , find ( a raised to 6 - 1/a raised to 6 )

Answer 1

★Given:-

$\sf a - \dfrac{1}{a}= \dfrac{8}{3}$

★To find:-

$\sf a^6 - \dfrac{1}{a^6}$

★Solution:-

First, we will find $\sf a^2 + \dfrac{1}{a^2}$

We will use the identity (a-b)² = a²+b² - 2ab to find it.

$\sf \Bigg ( a - \dfrac{1}{a} \Bigg )^2 = (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 - 2 \times a \times \dfrac{1}{a}$

$\sf \implies \Bigg ( \dfrac{8}{3} \Bigg )^2 = a^2 + \dfrac{1}{a^2} - 2$

$\sf \implies \dfrac{64}{9} = a^2 + \dfrac{1}{a^2} - 2$

$\sf \implies \dfrac{64}{9} + 2= a^2 + \dfrac{1}{a^2}$

$\sf \implies \dfrac{64+18}{9}= a^2 + \dfrac{1}{a^2}$

$\sf \implies \dfrac{82}{9}= a^2 + \dfrac{1}{a^2}$

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Now, we will find $\sf a + \dfrac{1}{a}$

We will use the identity (a+b)² = a² + b² + 2ab to find it.

$\sf \implies \Bigg (a+\dfrac{1}{a} \Bigg )^2= (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 + 2 \times a \times \dfrac{1}{a}$

$\sf \implies \Bigg (a+\dfrac{1}{a} \Bigg )^2= a^2 +\dfrac{1}{a^2} + 2$

$\sf \implies \Bigg (a+\dfrac{1}{a} \Bigg )^2= \dfrac{82}{9} + 2$

$\sf \implies \Bigg (a+\dfrac{1}{a} \Bigg )^2= \dfrac{82+18}{9}$

$\sf \implies \Bigg (a+\dfrac{1}{a} \Bigg )^2= \dfrac{100}{9}$

$\sf \implies a+\dfrac{1}{a}= \sqrt{\dfrac{100}{9}}$

$\sf \implies a+\dfrac{1}{a}= \dfrac{10}{3}$

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Now, finally we will find $\sf a^6 - \dfrac{1}{a^6}$

Here, we will use the identity,

$\boxed{ \sf a^6 - b^6 = (a-b)(a^2+b^2 +ab)(a+b)(a^2 + b^2 - ab)}$

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$\small \sf \implies a^6 - \dfrac{1}{a^6} = \Bigg ( a - \dfrac{1}{a} \Bigg ) \Bigg ( (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 + a \times \dfrac{1}{a} \Bigg ) \Bigg ( a + \dfrac{1}{a} \Bigg ) \Bigg ( (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 - a \times \dfrac{1}{a} \Bigg )$

$\small \sf \implies a^6 - \dfrac{1}{a^6} = \Bigg (\dfrac{8}{3} \Bigg ) \Bigg ( a^2 + \dfrac{1}{a^2} + 1 \Bigg ) \Bigg ( \dfrac{10}{3} \Bigg ) \Bigg ( a^2 + \dfrac{1}{a^2} - 1 \Bigg )$

$\small \sf \implies a^6 - \dfrac{1}{a^6} = \Bigg (\dfrac{8}{3} \Bigg ) \Bigg ( \dfrac{82}{9} + 1 \Bigg ) \Bigg ( \dfrac{10}{3} \Bigg ) \Bigg ( \dfrac{82}{9} - 1 \Bigg )$

$\small \sf \implies a^6 - \dfrac{1}{a^6} = \Bigg (\dfrac{8}{3} \Bigg ) \Bigg ( \dfrac{82+9}{9} \Bigg ) \Bigg ( \dfrac{10}{3} \Bigg ) \Bigg ( \dfrac{82-9}{9}\Bigg )$

$\small \sf \implies a^6 - \dfrac{1}{a^6} = \Bigg (\dfrac{8}{3} \Bigg ) \Bigg ( \dfrac{91}{9} \Bigg ) \Bigg ( \dfrac{10}{3} \Bigg ) \Bigg ( \dfrac{73}{9}\Bigg )$

$\sf \implies a^6 - \dfrac{1}{a^6} = \dfrac{531440}{729}$

Hope it helps