Question

a4+1/a4 =2 find a8+1/a8

Answer 1

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Answer 2

Correct Question :-

If $\sf a^4 + \dfrac{1}{a^4}$, the find $\sf a^8 + \dfrac{1}{a^8}$.

Answer :-

$\tt a^{8} + \dfrac{1}{a^8} =2$

Solution :-

$\sf a^4 + \dfrac{1}{a^4} = 2$

Squaring on both sides

$\sf \left( a^4 + \dfrac{1}{a^4} \right)^2 = {(2)}^2$

$\sf \left( a^4 + \dfrac{1}{a^4} \right)^2 = 4$

We know that

(x + y)² = x² + y² + 2xy

Here,

$\tt x = {a}^{4} \: and \:y = \dfrac{1}{ {a}^{4} }$

By substituting the values

$\sf {( {a}^{4} )}^{2} + \left( \dfrac{1}{a^4} \right)^2 + 2 \times {a}^{4} \times \dfrac{1}{ {a}^{4} } = 4$

$\sf {( {a}^{4} )}^{2} + \left( \dfrac{1}{a^4} \right)^2 + 2= 4$

$\sf a^{4(2)} + \dfrac{ {1}^{2} }{(a^4 )^{2} } + 2= 4$

$\sf a^{8} + \dfrac{1}{a^{4(2)} } + 2= 4$

$\sf a^{8} + \dfrac{1}{a^8} + 2= 4$

$\sf a^{8} + \dfrac{1}{a^8} =4 - 2$

$\sf a^{8} + \dfrac{1}{a^8} =2$

$\bf \therefore a^{8} + \dfrac{1}{a^8} =2$

Identity used :-

• (x + y)² = x² + y² + 2xy