If x=sqrt[3]{5}+2, then the value of x^{3}-6 | vedclass

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(D) Given $x = \sqrt[3]{5} + 2$.Subtracting $2$ from both sides,we get $x - 2 = \sqrt[3]{5}$.Now,cubing both sides of the equation:$(x - 2)^3 = (\sqrt

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(D) Given $x = \sqrt[3]{5} + 2$.Subtracting $2$ from both sides,we get $x - 2 = \sqrt[3]{5}$.Now,cubing both sides of the equation:$(x - 2)^3 = (\sqrt[3]{5})^3$Using the algebraic identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we expand the left side:$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = 5$$x^3 - 6x^2 + 12x - 8 = 5$To find the value of $x^3 - 6x^2 + 12x - 13$,we subtract $5$ from both sides of the equation:$x^3 - 6x^2 + 12x - 8 - 5 = 0$$x^3 - 6x^2 + 12x - 13 = 0$.

If $x=\sqrt[3]{5}+2,$ then the value of $x^{3}-6 x^{2}+12 x-13$ is

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