If x=2+2^{frac{2}{3}}+2^{frac{1}{3}},then x^3 | vedclass

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(B) Given $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$.Subtracting $2$ from both sides,we get $x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$.Cubing both sides,we have

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$2$

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(B) Given $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$.Subtracting $2$ from both sides,we get $x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$.Cubing both sides,we have $(x-2)^3 = (2^{\frac{2}{3}}+2^{\frac{1}{3}})^3$.Using the identity $(a+b)^3 = a^3+b^3+3ab(a+b)$,we get:$x^3-6x^2+12x-8 = (2^{\frac{2}{3}})^3 + (2^{\frac{1}{3}})^3 + 3(2^{\frac{2}{3}})(2^{\frac{1}{3}})(2^{\frac{2}{3}}+2^{\frac{1}{3}})$.$x^3-6x^2+12x-8 = 4 + 2 + 3(2^1)(x-2)$.$x^3-6x^2+12x-8 = 6 + 6(x-2)$.$x^3-6x^2+12x-8 = 6 + 6x - 12$.$x^3-6x^2+12x-8 = 6x - 6$.Rearranging the terms,we get $x^3-6x^2+6x = 8-6 = 2$.

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

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