Let $\mathrm{f}(x)=x^3+x-1$
A root of $\mathrm{f}(x)$ exists, if $\mathrm{f}(x)=0$ for at least one value of $x$.
$\begin{aligned}
& \mathrm{f}(0)=-1 < 0 \\
& \mathrm{f}(1)=1>0
\end{aligned}$
$\therefore \quad$ By intermediate value theorem, there has to be a point ' $c$ ' between 0 and 1 such that $\mathrm{f}(x)=0$.
$\therefore \quad$ The given equation has exactly one real root.
Alternate Method:
Let $\mathrm{f}(x)=x^3+x-1$
$\begin{array}{ll}
\therefore \quad \mathrm{f}^{\prime}(x)=3 x^2+1 \\
\quad \Rightarrow \mathrm{f}^{\prime}(x)>0 \quad \forall x \in \mathrm{R}
\end{array}$
$\Rightarrow \mathrm{f}(x)$ is an increasing function.
$\Rightarrow \mathrm{f}(x)$ intersects $\mathrm{X}$-axis at only one point.
$\therefore \quad$ The given equation has exactly one real root.