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

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

Question

(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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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

Similar Questions

$3 \div \left[ (8 - 5) \div \left\{ (4 - 2) \div \left( 2 + \frac{8}{13} \right) \right\} \right] = ?$

$\frac{69842 \times 69842 - 30158 \times 30158}{69842 - 30158} = ?$

The difference between $1 \frac{3}{16}$ and its reciprocal is:

If $\sqrt{3} = 1.732$,then the value of $\frac{9 + 2\sqrt{3}}{\sqrt{3}}$ is

Find the smallest fraction which when added to $\frac{2}{5} \times \frac{15}{21} \times \frac{7}{10} \times \frac{3}{8}$ gives a whole number.

Vedclass Test Series

Exam Paper Generator

Online Exam Module