Question

The number of real roots of the cubic equation $x^3-3x+1=0$ is

Answer 1

For the given cubic, we have:
$f^{\prime }\left(x\right)=3(x^2-1)$
$\Rightarrow$Roots of $f^{\prime }\left(x\right)=0$ are $-1$ and $+1$
$f\left(1\right)=1-3+1=-1$
$f(-1)=-1+3+1=3$
Clearly $f\left(1\right)\cdot f(-1)<0$
$\Rightarrow$ The given cubic equation has $3$ real roots.