Question

Show that $x^3-3x^2+3x-1$ is exactly divisible by $(x-1)$.

Answer 1

Let $f\left(x\right)=x^3-3x^2+3x-1$.

We find the remainder using, $x-1=0⟹x=1$.

Then, $f\left(1\right)=(1{)}^3-3(1{)}^2+3\left(1\right)-1$ $=1-3+3-1=0$.

According to the remainder theorem, we have $f\left(x\right)$ is exactly divisible by $(x-1)$ as the remainder is $0$ when divided by $(x-1)$.

Hence, showed.