Question

If $a,b,c$ are the roots of $x^3-3x^2+3x+26=0$ and $w$ is cube roots of unify then the value of $\frac{a-1}{b-1}+\frac{b-1}{c-1}+\frac{c-1}{a-1}=$

• A. $-3w^2$
• B. $3w^2$
• C. $w^2$
• D. $2w^2$

Answer 1

The correct option is B $3w^2$

The equation given $x^3-3x^2+3x+26=0$ can be written as,

$x^3-3x^2+3x-1+27=0$

or, $(x-1{)}^3=(-3{)}^3$

or, $x=1-3,1-3\omega ,1-3{\omega }^2$. [ Where ${\omega }^3=1$]

Let, $a=-2,b=1-3\omega ,c=1-3{\omega }^2$

or, $a-1=-3,b-1=-3\omega ,c-1=-3{\omega }^2$.......(1).

Now,

$\frac{a-1}{b-1}+\frac{b-1}{c-1}+\frac{c-1}{a-1}$

$=\frac{-3}{-3\omega }+\frac{-3\omega }{-3{\omega }^2}+\frac{-3{\omega }^2}{-3}$ [ Using (1)]

$={\omega }^2+{\omega }^2+{\omega }^2$

$=3{\omega }^2$.