* 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

**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 .**

*f(x)*
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