If x = 2 + 2^{2/3} + 2^{1/3},then x^3 - 6x^2 | vedclass

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(B) Given $x = 2 + 2^{2/3} + 2^{1/3}$.Subtracting $2$ from both sides,we get $x - 2 = 2^{2/3} + 2^{1/3}$.Cubing both sides,we use the identity $(a + b

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$2$

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(B) Given $x = 2 + 2^{2/3} + 2^{1/3}$.Subtracting $2$ from both sides,we get $x - 2 = 2^{2/3} + 2^{1/3}$.Cubing both sides,we use the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.$(x - 2)^3 = (2^{2/3})^3 + (2^{1/3})^3 + 3(2^{2/3})(2^{1/3})(2^{2/3} + 2^{1/3})$.$(x - 2)^3 = 2^2 + 2^1 + 3(2^1)(x - 2)$.$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = 4 + 2 + 6(x - 2)$.$x^3 - 6x^2 + 12x - 8 = 6 + 6x - 12$.$x^3 - 6x^2 + 12x - 8 = 6x - 6$.Rearranging the terms to find $x^3 - 6x^2 + 6x$,we get:$x^3 - 6x^2 + 6x = 8 - 12 + 6 = 2$.Thus,the correct option is $B$.

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

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