Cubic equation in three variables

Consider a polynomial $P(t) =t^{3} + at^{2} + bt + c$ with roots x, y and z

Since x + y + z = 4 ,it follows that a = - 4 .

We have $x^{2} + y^{2} + z^{2} = (x + y + z) ^{2} - 2 (xy+yz+zx)$

It follows that $b = xy + yz + z = 1$

Since x, y, z are roots of P(t),

• $x^{3} - 4x^{2} + x + c = 0$
• $y^{3} - 4y^{2} + y + c = 0$
• $z^{3} - 4z^{2} + z + c = 0$

Adding these equalities and using equations of system, we obtain c = 6

Hence, $P(t) = t^{3} - 4t^{2} + t + 6 = (t+1)(t^{2} - 5t +6) = (t+1)(t-2)(t-3)$

So the roots of P(t) are -1,2,3 and thus x, y, z = ordered pairs of (-1,2,3) and S = - 1+2+3 =4.

Let x,y,z be the root of $p(t) =t^3 +at^2+bt+c$ so, all values of x are $x,y,z$ from the given information, $x+y+z= \boxed{4}$

The first equation tells you that the sum of x, y, and z is 4, so S = 4