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 ^3-6 t ^2+8 t -2=0 $
$\therefore \text { Product of roots }=2$