Question

If $a,b,c$ are three distinct non zero real numbers satisfying the equations $\frac{1}{a-1}+\frac{1}{a-2}+\frac{1}{a-3}=1,\frac{1}{b-1}+\frac{1}{b-2}+\frac{1}{b-3}=1$ and $\frac{1}{c-1}+\frac{1}{c-2}+\frac{1}{c-3}=1$ then $(a-1)(b-1)(c-1)$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Clearly $a,b,c$ are roots of the equation $\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}=1$ replace $x$ by $t+1$
$\frac{1}{t}+\frac{1}{t-1}+\frac{1}{t-2}=1$
Clearly, equation will have roots as a $-1,b-1,c-1$
$\therefore (t-1)(t-2)+t(t-2)+t(t-1)=t(t-1)(t-2)$
$\therefore t^3-6t^2+8t-2=0$
$\therefore \text{ Product of roots }=2$