Question

If $Î\pm ,β$ and $γ$ are the roots of the equation $x^3−3x^2+3x+7=0$, and w is cube root of unity, then the value of $\frac{Î\pm −1}{β−1}+\frac{β−1}{γ−1}+\frac{γ−1}{Î\pm −1}$ is equal to

• A. $3w^2$
• B. $3/w$
• C. $2w^2$
• D. none of these

Answer 1

Answer: (A)

We have,
$x^3−3x^2+3x+7=0$
$\left(x+1\right)\left(x^2−4x+7\right)=0$
$x+1=0$ or $x^2−4x+7=0$
$x=−1$ or $x=\frac{4Â\pm \sqrt{16−28}}{2}$
$⇒x=−1$ or $x=\frac{4Â\pm 2\sqrt{3}i}{2}$
$⇒x=−1$ or $x=2Â\pm \sqrt{3}i$
$⇒x=−1$ or $x=1−2w$, $1−2w,1−2w^2$
$\left[2w=−1+\sqrt{3}i,2w^2=−1−\sqrt{3}i\right]$
$∴Î\pm =−1,β=1−2w$ and $γ=1−2w^2$
Now, $\frac{Î\pm −1}{β−1}+\frac{β−1}{γ−1}+\frac{γ−1}{Î\pm −1}$
$=\frac{−1−1}{1−2w−1}+\frac{1−2w−1}{1−2w^2−1}=\frac{1−2w^2−1}{−1−1}$
$=\frac{2}{2w}+\frac{2w}{2w^2}+\frac{2w^2}{2}$
$=\frac{1}{w}+\frac{1}{w}+w^2=\frac{2}{w}+w^2$
$=2w^2+w^2=3w^2$