Question

If $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$ then prove that $x^3-6x^2+6x-2=0$

Answer 1

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

Cubing both sides we get

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

$x^3-8-3\times 2x(x-2)=2+4+3\times 2^{\frac{2}{3}}\times 2^{\frac{1}{3}}(2^{\frac{2}{3}}+2^{\frac{1}{3}})$ using ${(a+b)}^3=a^3+b^3+3ab(a+b)$

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

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

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

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