Let f(x)=x3+x−1 A root of f(x) exists, if f(x)=0 for at least one value of x. ‌f(0)=−1<0 ‌f(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.