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Problem:Let a,b,c be distinct real numbers.Then the number of real solution of $(x-a)^3+(x-b)^3+(x-c)^3=0$ is

(A) 1

(B) 2

(C) 3

(D) depends on a,b,c

Solution: Ans: (A)

Let $f(x)=(x-a)^3+(x-b)^3+(x-c)^3$

$=> f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2=0$ $=> f'(x)=(x-a)^2+(x-b)^2+(x-c)^2=0$ $=> x=a, x=b, x=c$

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 .