Try this beautiful problem from TOMATO Objective no. 258 based on Real Roots of a Cubic Polynomial.
Problem: Real Roots of a Cubic Polynomial
Let a,b,c be distinct real numbers. Then the number of real solution of [latex](x-a)^3+(x-b)^3+(x-c)^3=0[/latex] is
(A) 1
(B) 2
(C) 3
(D) depends on a,b,c
Solution: Ans: (A)
Let [latex]f(x)=(x-a)^3+(x-b)^3+(x-c)^3[/latex]
[latex]=> f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2=0[/latex]
[latex]=> f'(x)=(x-a)^2+(x-b)^2+(x-c)^2=0[/latex]
[latex]=> x=a, x=b, x=c [/latex]
But it is not possible a quadratic equation has three roots.so, it implies that f'(x) has no real roots.But f(x) is a cubic polynomial. And we know a cubic polynomial must have at least one real root ( we know all polynomial curves are continuous ,so it cuts either of the axes at least once).So, number of real root of the given equation is 1 .
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