Let f(x)=x3+x−1A root of f(x) exists, if f(x)=0 for at least one value of x.f(0)=−1<0f(1)=1>0∴ By intermediate value theorem, there has to be a point ' c ' between 0 and 1 such that f(x)=0.∴ The given equation has exactly one real root.Alternate Method: Let f(x)=x3+x−1∴f′(x)=3x2+1⇒f′(x)>0∀x∈R⇒f(x) is an increasing function.⇒f(x) intersects X-axis at only one point.∴ The given equation has exactly one real root.