The equation \(x^3+x-1=0\) has

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