If ω is the cube root of unity (a+b ω+c ω^2)/(c+a ω+b ω^2)+(a+b ω+c ω^2)/(b+c ω+a ω^2)=

Correct answer: (D) -1. (a+b ω+c ω^2)/(c+a ω+b ω^2)+(a+b ω+c ω^2)/(b+c ω+a ω^2) (a+b ω+c ω^2)/(c+a ω+b ω^2)+(ω^2 (a+b ω+c ω^2 ))/(b ω^2+c+a ω) = (1+ω^2 ) ((a+b ω+c ω^2)/(c+a ω+b ω^2) )=-((a ω+b ω^2+c ))/((c+a ω+b ω^2 ))=-1

Q49 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

If $\omega$ is the cube root of unity $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}=$

A
2
B
-2
C
1
Correct answer
D
-1

Solution

$\begin{aligned} & \text { } \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} \\ & \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{\omega^2\left(a+b \omega+c \omega^2\right)}{b \omega^2+c+a \omega} \\ & =\left(1+\omega^2\right)\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)=-\frac{\left(a \omega+b \omega^2+c\right)}{\left(c+a \omega+b \omega^2\right)}=-1\end{aligned}$

Hence, the correct answer is option: 'D'.

If $\omega$ is the cube root of unity $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}=$

Solution

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