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 $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B\alpha+C\alpha^2$,where $A, B, C \geq 0$,then $5(3A-2B-C)$ is equal to:

If $i{z^4} + 1 = 0$,then $z$ can take the value

If $\alpha \neq 1$ is any $n^{th}$ root of unity,then $S = 1 + 3\alpha + 5\alpha^2 + \dots$ up to $n$ terms,is equal to

If $\omega$ is a complex cube root of unity and $(1+\omega)^7=A+B \omega$,then the values of $A$ and $B$ are,respectively.

${\left( {\frac{{1 + \sin \theta + i\cos \theta }}{{1 + \sin \theta - i\cos \theta }}} \right)^n} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module