Question

Let a, b, c be complex such that $ab+bc+ca=\frac{1}{2},(a+b+c)=2,abc=4$ then the value of $\frac{1}{ab+c-1}+\frac{1}{bc+a-1}+\frac{1}{ac+b-1}$ is

• A. $-\frac{2}{9}$
• B. $\frac{1}{9}$
• C. $\frac{2}{9}$
• D. $-\frac{1}{9}$

Answer 1

The correct option is A

$-\frac{2}{9}$

$ab+bc+ca=\frac{1}{2};a+b+c=2;abc=4$

$\frac{1}{ab+c-1}+\frac{1}{bc+a-1}+\frac{1}{ac+b-1}=?$

$\frac{1}{ab+c-1}=\frac{1}{ab+[2-a-b]-1}$ $(\because a+b+c=2$

$=\frac{1}{ab-a-b+1}$

$=\frac{1}{(a-1)(b-1)}$

$\therefore \frac{1}{ab+c-1}+\frac{1}{bc+a-1}+\frac{1}{ac+b-1}$

$=\frac{1}{(a-1)(b-1)}+\frac{1}{(c-1)(b-1)}+\frac{1}{(a-1)(c-1)}$

$=\frac{c-1+b-1+a-1}{(a-1)(b-1)(c-1)}$

$=\frac{\sum a-3}{\sum abc-\sum ab+\sum a-1}$

$=\frac{2-3}{4-\frac{1}{2}+2-1}$

$=-\frac{2}{9}$