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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 have $(x - 2)^3 = (2^{2/

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 have $(x - 2)^3 = (2^{2/3} + 2^{1/3})^3$.Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$,we get:$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = (2^{2/3})^3 + (2^{1/3})^3 + 3(2^{2/3})(2^{1/3})(2^{2/3} + 2^{1/3})$.$x^3 - 6x^2 + 12x - 8 = 4 + 2 + 3(2^1)(x - 2)$.$x^3 - 6x^2 + 12x - 8 = 6 + 6x - 12$.$x^3 - 6x^2 + 12x - 8 = 6x - 6$.Rearranging the terms,$x^3 - 6x^2 + 6x = 8 - 6 = 2$.

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

Similar Questions

If for real values of $x$,$\cos \theta = x + \frac{1}{x}$,then

For the equation $|x^2| + |x| - 6 = 0$,the roots are

If for some $p, q, r \in \mathbb{R}$,not all have the same sign,and one of the roots of the equation $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$ is also a root of the equation $x^{2}+2x-8=0$,then $\frac{q^{2}+r^{2}}{p^{2}}$ is equal to-

The roots of the equation $x^4 - 8x^2 - 9 = 0$ are

The number of real solutions of the equation $|x^2 + 4x + 3| + 2x + 5 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module