If $\frac{K}{k^2-2k+1}=\frac{1}{6}$, then what is the value of $\frac{1}{k^3}+k^3$ ?

492

If x + \(\frac{1}{x}\)  = n

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

If $\frac{K}{k^2-2k+1}=\frac{1}{6}$

Then, $\frac{1}{k^3}+k^3$ = ?

$k^2$-2k+1 = 6k

= $k^2$-8k+1 = 0

Divide the equation by k on both sides,

k + \(\frac{1}{k}\) = 8

$x^3 +\frac{1}{x^3}$ = 8³ - 3 × 8

$x^3 +\frac{1}{x^3}$ = 512 - 24 = 492