Question

${x}^{2} + \frac{1}{ {x}^{2} } = 102 \: find \: the \: value \: of \: {x}^{3} + \frac{1}{ {x}^{3} }$
I will mark the best answer as the brainliest

Answer 1

Given x^2 + (1/x^2) = 102.

It can be written as :

= > x^2 + (1/x^2) + 2 = 104

= > (x + 1/x)^2 = 104

$= > x + \frac{1}{x} = \sqrt{104}$

$= > x + \frac{1}{x} = 2\sqrt{26}$

On cubing both sides, we get

$= > (x + \frac{1}{x})^3 = (2\sqrt{26} )^3$

$= > x^3 + \frac{1}{x^3} + 3 * x * \frac{1}{x}(x + \frac{1}{x} ) = 208\sqrt{26}$

$= > x^3 + \frac{1}{x^3} + 3(2\sqrt{26}) = 208\sqrt{26}$

$= > x^3 + \frac{1}{x^3} = 208\sqrt{28} - 6\sqrt{26}$

$= > x^3 + \frac{1}{x^3} = 202\sqrt{26}$

Hope this helps!

Answer 2

I Hope My Answer Will Be Right.