Question

$x=2+2^{1/3}+2^{2/3}$. find the value of $x^3-6x^2+6x$.

Answer 1

Consider $x=2+2^{1/3}+2^{2/3}$

$\Rightarrow x-2=2^{\frac{1}{3}}+2^{\frac{2}{3}}...\left(1\right)$

Cubing on both sides we get

$\Rightarrow (x-2{)}^3={(2^{\frac{1}{3}}+2^{\frac{2}{3}})}^3$

$\Rightarrow x^3-8+12x-6x^2={\left(2^{\frac{1}{3}}\right)}^3{(1+2^{\frac{1}{3}})}^3$

$\Rightarrow x^3-8+12x-6x^2=2(1+2+3(2^{1/3})+3(2^{2/3}\left)\right)$

$\Rightarrow x^3-8+12x-6x^2=2(1+2+3(2^{1/3}+2^{2/3}\left)\right)$

$\Rightarrow x^3-8+12x-6x^2=2(1+2+3(x-2\left)\right)\left[by\right(1\left)\right]$

$\Rightarrow x^3-8+12x-6x^2=2(3+3(x-2\left)\right)$

$\Rightarrow x^3-8+12x-6x^2=6+6x-12$

$\Rightarrow x^3-8+12x-6x^2=6x-6$

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

$\Rightarrow x^3-6x^2+6x-2=0$

$\therefore x^3-6x^2+6x=2$