Complex Numbers And Quadratic Equations question 157
Question: If the cube roots of unity be $ 1,\omega ,{{\omega }^{2}}, $ then the roots of the equation $ {{(x-1)}^{3}}+8=0 $ are [IIT 1979; MNR 1986; DCE 2000; AIEEE 2005]
Options:
A) $ -1,1+2\omega ,1+2{{\omega }^{2}} $
B) $ -1,1-2\omega ,1-2{{\omega }^{2}} $
C) $ -1,-1,-1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{(x-1)}^{3}}=-8\Rightarrow x-1={{(-8)}^{1/3}} $
Þ $ x-1=-2,-2\omega ,-2{{\omega }^{2}} $
Þ $ x=-1,1-2\omega ,1-2{{\omega }^{2}} $ Trick: By inspection, we see that (b) satisfies the equation i.e, $ {{(-1-1)}^{3}}+8=0,{{(1-2\omega -1)}^{3}}+8=0 $ $ {{(1-2{{\omega }^{2}}-1)}^{3}}+8=0 $ .
BETA
