Question

x^3 + 1/x^3 = 488 then x^3 - 1/x^3 =

Answer 1

Answer:

$\large{\underline{\boxed{\sf x {}^{3} - \frac{1}{x ^{3} } = 57 \sqrt{60}}}}$

Step-by-step explanation:

Given that ;

$\tt x^{3} + \dfrac{1}{x {}^{3} } = 488$

We know that ;

$\tt (x + \dfrac{1}{x} ) {}^{3} = x {}^{3} + \dfrac{1}{x {}^{3} } + 3(x + \dfrac{1}{x} )$

Let the value of (x + 1/x) be a.

$\implies \tt a {}^{3} = 488 + 3a \\ \\ \implies \tt a {}^{3} - 3a - 488 = 0$

Using Hit-and-Trial method, put a = 8.

$\implies \tt (8) {}^{3} - 3(8) - 488 \\ \\ \implies \tt 512 - 24 - 488 \\ \\ \implies \tt 512 - 512 \\ \\ \implies \tt 0$

Therefore, value of a is 8.

$\implies \tt x + \frac{1}{x} = a \\ \\ \implies \tt x + \frac{1}{x} = 8$

_______________________

We know that ;

$\tt (x - \frac{1}{x} ) {}^{2} = (x + \frac{1}{x} ) {}^{2} - 4 \\ \\ \implies \tt (x - \frac{1}{x} ) {}^{2} =( 8)^{2} - 4 \\ \\ \implies \tt (x - \frac{1}{x} ) {}^{2} = 64 - 4 \\ \\ \implies \tt (x - \frac{1}{x} ) {}^{2} = 60 \\ \\ \implies \tt x - \frac{1}{x} = \pm \sqrt{60}$

On cubing both sides,

$\implies \tt (x - \frac{1}{x} ) {}^{3} = ( \sqrt{60} ) {}^{3} \\ \\ \implies \tt x {}^{3} - \frac{1}{x {}^{3} } - 3(x - \frac{1}{x} ) = 60 \sqrt{60} \\ \\ \implies \tt x {}^{3} - \frac{1}{x {}^{3} } - 3( \sqrt{60} ) = 60 \sqrt{60} \\ \\ \implies \tt x {}^{3} - \frac{1}{x { }^{3} } = 60 \sqrt{60} - 3 \sqrt{60} \\ \\ \boxed{ \bf \therefore \: x {}^{3} - \frac{1}{x {}^{3} } = 57 \sqrt{60} }$