My question is about a step in my solution to this system of equations. $$ \left\{ \begin{array}{c} a+b+c=0 \\ a^3+b^3+c^3 = 90\\ a^7+b^7+c^7 = 75810 \end{array} \right. $$ My solution: \begin{eqnarray*} \because a+b+c = 0 \\ \therefore a^3+b^3+c^3 = 3abc, abc = 30 \\ a^7+b^7+c^7 = -7ab(a+b)(a^2+ab+b^2)^2 = 7abc(a^2+ab+b^2)^2 \\ \therefore (a^2+ab+b^2)^2 = 361 \\ a^2+ab+b^2 = \pm 19 \\ a^3 + a^2b + b^2a = 19a \\ a^3 + ab(a+b) = 19a \\ a^3 -19a -30 = 0 \\ (a-5)(a+3)(a+2) = 0, a \in \{5,-3,-2\} \end{eqnarray*}$
Now I considered the case when $a^2+ab+b^2 = 19$, It provided the right answer to the system, yet when considering the negative case ($a^2+ab+b^2 = -19$), It doesn't provide the right answer when solving the cubic equation formed after multiplying by $a$. Why does this happen?
$a+b+c = 0$
$a^3+b^3+c^3 = 90$
$a^7+b^7+c^7 = 75810$
Clearly we can easily conclude that $c = -a-b$
$a^3+b^3+c^3 = (a+b+c)^3-3*(a+b)*(a+c)*(b+c)$
$90 = -3*(a+b)*(a+c)*(b+c)$
substitute that $c = -a-b$
$90 = -3*a*b*(a+b)$
Therefore $a*b*(a+b) = -30$
$a^7+b^7+c^7 = (a+b+c)^7-7*(b+c)*(a+c)*(a+b)*(a^4+2*a^3*b+2*a^3*c+3*a^2*b^2+5*a^2*b*c+3*a^2*c^2+2*a*b^3+5*a*b^2*c+5*a*b*c^2+2*a*c^3+b^4+2*b^3*c+3*b^2*c^2+2*b*c^3+c^4)$
substitute that $c = -a-b$
$a^7+b^7+c^7 = -7*b*a*(a+b)*(a^2+a*b+b^2)^2$
$75810 = -7×-30*(a^2+a*b+b^2)^2$
$361 = (a^2+a*b+b^2)^2$
Now $a^2+a*b+b^2 = \sqrt(361)$
$a^2+a*b+b^2 = ±19$
But only the positive value will work
Proof:
Since $a+b+c = 0$, then either $a$ or $b$ or $c$ must be negative, they can't be all positive.
Since we saw that $a*b*(a+b) = -30$ It means clearly that either $a$ or $b$ must be negative
$a^2+a*b+b^2 = \sqrt(361)$
$(a+b)^2-2*a*b+a*b = \sqrt(361)$
$(a+b)^2-a*b = \sqrt(361)$
Since we've proved that either $a$ or $b$ must be negative, and we know that the square of any number is always positive ( Checking ) $(positive) - (negative)$ must be $(positive)$
$a^2+a*b+b^2 = 19$
$a^3+a^2*b+a*b^2 = 19*a$
$a^3+a*b(a+b) = 19*a$
$a^3-30 = 19*a$
$a^3-19*a-30 = 0$
Which factors to $(a-5)*(a+2)*(a+3)$
notice that a and b are symmetry in all the equations, therefore they will satisfy the same polynomial.
So the equation for b would also factor to $(b-5)*(b+2)*(b+3)$
The results of $a$ or $b$ or $c$ are $5$ or $-2$ or $-3$