Search any question & find its solution
Question:
Answered & Verified by Expert
If the cube roots of unity are $1, \omega, \omega^2$ then the roots of the equation $(x-1)^3+8=0$, are
Options:
Solution:
1344 Upvotes
Verified Answer
The correct answer is:
$-1,1-2 \omega, 1-2 \omega^2$
$-1,1-2 \omega, 1-2 \omega^2$
$(x-1)^3+8=0 \Rightarrow(x-1)=(-2)(1)^{1 / 3}$ $\Rightarrow x-1=-2$ or $-2 \omega$ or $-2 \omega^2$ or $n=-1$ or $1-2 \omega$ or $1-2 \omega^2$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.