Polynomials — Maths STD 9 — Question

The given equation is
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
This can be written as
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
$=x^3-3 x^2+3 x-7=a x^3+b x^2+c x+a x^2+b x+c$
$=x^3-3 x^2+3 x-7=a x^3+(a+b) x^2+(b+c) x+c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$
Hence, correct option is $(c).$