Question

If $x=2+2^{\frac{1}{3}}+2^{\frac{2}{3}}$, then the values of $x^3-6x^2+6x=$

Answer 1

Given that, $x=2+2^{\frac{1}{3}}+2^{\frac{2}{3}}$ ……. (1)

Let, $2^{\frac{1}{3}}=t$, then, $t^3=2$

By equation (1) and we get,

$x=2+t+t^2$

$x-2=t+t^2$

Cubing both sides, and we get,

${(x-2)}^3={(t+t^2)}^3$

$=t^3+t^6+3t^4+3t^5$

Put $t^3=2$ and we get,

$=2+{\left(2\right)}^2+3\left(2\right)t+3\left(2\right)t^2$

$=6t^2+6t+6$

$=6(t^2+t)+6$

$=6(x-2)+6$

${(x-2)}^3=6(x-2)+6$

${(x-2)}^3-6(x-2)-6=0$

$x^3-8-6x^2+12x-6x+12-6=0$

$x^3-6x^2-6x-2=0$

$x^3-6x^3-6x^2=2$

Hence, this is the answer.