Complex Numbers And Quadratic Equations question 236
Question: $ {{(27)}^{1/3}}= $
Options:
A) 3
B) $ 3,3i,3i^{2} $
C) $ 3,3\omega ,3{{\omega }^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ x={27^{1/3}} $ $ =x^{3}-27=0 $
Þ $ (x-3)(x^{2}+3x+9)=0 $ $ x=3,x=3( \frac{-1\pm i\sqrt{3}}{2} ) $ . Hence, roots are $ 3,3\omega ,3{{\omega }^{2}} $ . Trick: As we know $ {{(27)}^{1/3}} $ must have 3 roots, so (a) option cannot be the best. Here (c) satisfies.
BETA
