The equation

          (x-a)³+(x-b)³+(x-c)³=0  ,  has

(A) all the roots real

(B) one real and two imaginary roots

(C) three real roots namely x=a,x=b,x=c

(D) none 

is d answer A................... ??

no it's (b)

 someone pls answer........................

Damn this is a crazy Q

It must have some conditions on a,b,c as if a=b=c then it has three real roots !!

Assuming that a!=b!=c I can prove that the eqn can't have 3 real roots and hence a,b, c being real and complex roots occuring in conjugate pairs the option (b) is correct .

The proof goes as follows :

Denote the LHS by f(x) and assume that it has three real roots . Then f'(x) =0must have atleast two real roots

That is (x-a)^2 + (x-b)^2 + (x-c)^2 = 0 has atleast two real roots

but it is give that a!=b!=c , so the above eqn has no real roots at all !!

Hence it must have one real and two conjugate imaginary roots .

Yeah, pls check if there is any condition on a,b and c