The value of (8)^{1/3} is | vedclass

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

Question

(D) Let $(8)^{1/3} = x$. Then $x^3 = 8$,which implies $x^3 - 8 = 0$. Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we get $(x - 2

Vedclass · Accepted Answer

All of these

Vedclass · Answer

(D) Let $(8)^{1/3} = x$. Then $x^3 = 8$,which implies $x^3 - 8 = 0$. Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we get $(x - 2)(x^2 + 2x + 4) = 0$. This gives $x = 2$ or $x^2 + 2x + 4 = 0$. Solving the quadratic equation $x^2 + 2x + 4 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}$. Thus,the cube roots of $8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$. Therefore,all the given options are correct.

The value of $(8)^{1/3}$ is

Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}=$

The product of all the values of $(\sqrt{3}-i)^{\frac{3}{7}}$ is

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$, where $i = \sqrt{-1}$, then $(z^{201} - i)^{8}$ is equal to:

If $\omega$ is a complex cube root of unity,then $(x - y)(x\omega - y)(x\omega^2 - y) = $

If $z^2+z+1=0$,then find the value of $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$,where $z$ is a complex cube root of unity.

Vedclass Test Series

Exam Paper Generator

Online Exam Module