If $x^4 - 142 x^2 + 1 = 0$, then the value of $x^3 + \frac{1}{x^3}$ is :

1692

We know that,

x² + \(\frac{1}{x^2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

We also know that,

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

If $x^4 - 142 x^2 + 1 = 0$,

then the value of $x^3 + \frac{1}{x^3}$ = ?

If $x^4 - 142 x^2 + 1 = 0$,

Dividing both sides by x² of the above equation,

x² + \(\frac{1}{x^2}\) = 142

then, x + \(\frac{1}{x}\) = \(\sqrt {142 + 2}\) = 12

then the value of $x^3 + \frac{1}{x^3}$ = 12³ - 3 × 12 = 1728 - 36 = 1692