Question

Let a, b and c be the roots of $x^3-x+1=0$, then the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$ equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$-2$

Given that $a,b$ and $c$ are the roots of equation $x^3-x+1=0$

Therefore,

$A=1,B=0,C=-1,D=1$

Now,

Sum of roots $=\frac{-B}{A}$

$\Rightarrow a+b+c=0$

Sum of product of roots $=\frac{C}{A}$

$\Rightarrow ab+bc+ca=\frac{-1}{1}=-1$

Product of roots $=\frac{-D}{A}$

$\Rightarrow abc=\frac{-1}{1}=-1$

Now,

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

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

$=\frac{(b+c+bc+1)+(a+c+ac+1)+(a+b+ab+1)}{abc+(a+b+c)+(ab+bc+ca)+1}$

$=\frac{2(a+b+c)+(ab+bc+ca)+3}{abc+(a+b+c)+(ab+bc+ca)+1}$

$=\frac{2\left(0\right)+(-1)+3}{(-1)+0+(-1)+1}$

$=\frac{3-1}{-1}$

$=-2$

Thus the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$ is $-2$.

Hence the correct answer is $\left(D\right)-2$