Question

$\tt\pink{4b^2+\dfrac{1}{b^2}=2} \\ \sf\blue{8b^3+\dfrac{1}{b^3}}=\red?$​

Answer 1

Questions :-

$\: \tt\pink{4b^2+\dfrac{1}{b^2}=2} \\ \: \sf\blue{8b^3+\dfrac{1}{b^3}=}\red?$

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Solution :-

Given Information :-

$\: \tt\pink{4b^2+\dfrac{1}{b^2}=2}$

To Find :-

$\: \sf\blue{8b^3+\dfrac{1}{b^3}=}\red?$

Calculation :-

$\sf : \implies (2b + \frac{1}{b} )^{2} = {4b}^{2} + \frac{1}{ {b}^{2} } + 4$

$\sf : \implies {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\sf : \implies (2b + \frac{1}{b} )^{2} = 6$

$\sf : \implies (2b + \frac{1}{b} ) = \red{ ±\sqrt{6} }$

$\sf : \implies(a + b)^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

$\sf : \implies(2b + \frac{1}{b} )^{2} = {8b}^{3} + { \frac{1}{b ^{3} }} + 6(2b + \frac{1}{b} )$

CASE 1 :-

Substituting the value of

$\sf : \implies (2b + \frac{1}{b} ) = \sqrt{6}$

$\sf : \implies { (\sqrt{6} )}^{3} = (8b)^{3} + \frac{1}{b^{3} } + 6( \sqrt{6} )$

$\sf : \implies 6 \sqrt{6} = {8b}^{3} + \frac{1}{b^{3} } + 6 \sqrt{6}$

$\sf : \implies {8b}^{3} + \frac{1}{b^{3} } = 0$

CASE 2 :-

Even, if you take the value of,

$\sf : \implies(2b + \frac{1}{b} ) \: as \: - \sqrt{6}$

$\sf : \implies - 6 \sqrt{6} = 8b^{3} + \frac{1}{b^{3} } - 6 \sqrt{6}$

$\sf : \implies {8b}^{3} + \frac{1}{ {b}^{3} } = 0$

In both the cases, the answer will be 0.

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Final Answer :-

$\sf \red{± \sqrt{6} }$

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Note :-

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Answer 2

Answer:

square 6 is your answer thank your