Question

solve the above question​

Answer 1

Given

$\sf \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 83$

To find

$\sf \to \: {x}^{3} - \dfrac{1}{ {x}^{3} }$

So Now Take

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x}$

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{2} = 83 - 2$

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{2} = 81$

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{} = \sqrt{ 81}$

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{} = 9$

Now Take

$\sf \to \: \bigg(x - \dfrac{1}{x} \bigg) ^{3} = {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \times x \times \dfrac{1}{x} \bigg(x - \dfrac{1}{x} \bigg)$

$\sf \to \: \bigg(x - \dfrac{1}{x} \bigg) ^{3} = {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \bigg(x - \dfrac{1}{x} \bigg)$

Now put the value

$\sf \to \bigg(x - \dfrac{1}{x} \bigg) ^{} = 9$

We get

$\sf \to \: (9) {}^{3} = {x}^{3} - \dfrac{1}{ {x}^{3} } - 3(9)$

$\sf \to \: (729 ) = {x}^{3} - \dfrac{1}{ {x}^{3} } - 27$

$\sf \to \: {x}^{3} - \dfrac{1}{ {x}^{3} } = 729 + 27$

$\sf \to \: {x}^{3} - \dfrac{1}{ {x}^{3} } = 756$

Answer

$\sf \to \: {x}^{3} - \dfrac{1}{ {x}^{3} } = 756$