If $\alpha, \beta$ and $\gamma$ are the roots of $x^3 + 8 = 0$,then the equation whose roots are $\alpha^2, \beta^2$ and $\gamma^2$ is
- A$x^3 - 8 = 0$
- B$x^3 - 16 = 0$
- C$x^3 + 64 = 0$
- D$x^3 - 64 = 0$
Explore More
Similar Questions
For $n \in N$,if $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right)$,then $(A_1 A_2 A_3 A_4)^4 =$
If $(\sqrt{3}-i)^{n}=2^{n}, n \in N$,then the least possible value of $n$ is
If $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$,then $5 \alpha^4 + 5 \beta^4 + 7 \alpha^{-1} \beta^{-1}$ is equal to
The roots of the equation $x^4 - 1 = 0$ are
Easy
View Solution$(1+\sqrt{3}i)^6-(\sqrt{3}+i)^6=$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo