Question

please please please read this question and answer

Answer 1

Given that :

$x + \frac{1}{x} = 4$

As we know that :

${a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b) \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x}) }^{3} - 3(x + \frac{1}{x} ) \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = {4}^{3} - 3 \times 4 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 52$

So, the answer will be 52