An algebra problem by Puneet Pinku

By rearranging the equations, we get:

$a^3 + ax = - y$ $b^3 + bx = - y$ $c^3 + cx = - y$

From these, we can get the following 3 new equations:

(1) $a^3 + ax = b^3 + bx$

(2) $a^3 + ax = c^3 + cx$

(3) $b^3 + bx = c^3 + cx$

Equation (1) can be written in the following form:

$a^3 - b^3 = - (a - b)x$

And since a and b are distinct real numbers ( a - b ≠ 0 ), we can divide by (a-b) :

(4) $a^2 + ab + b^2 = - x$

Similarly, from (2) and (3):

(5) $a^2 + ac + c^2 = -x$

(6) $b^2 + bc + c^2 = -x$

By taking away (5) from (4):

      b^2 - c^2 + ab - ac = 0  

      b^2 - c^2 + ab - ac = 0 

      (b + c)(b - c) + a(b - c) = 0

(7) $(a + b + c)(b - c) = 0$

Now, due to the Zero Factor Theorem, the LHS of (7) is 0 if and only if at least one of its factors is 0. Since b and c are distinct real numbers, (b - c) ≠ 0 and therefore:

$\boxed {a + b + c = 0}$

Consider a polynomial of the form $px^3+px+y = 0$ . Clearly, the a,b,c satify the condition of being roots of the given equation, and hence by vieta roots, (-b/a) = a+b+c =0.(Here -b/a represents the sum of roots of the equation $ax^3+bx^2+cx+d$ .

Nice question. I also solved it using the same method. This question came in kv pre-jmo. How much are you getting in pre jmo? Will you be selected for jmo?