Question 1108050
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In a polynomial of degree n, the sum of the roots is -b/a, where a is the leading coefficient and b is the coefficient of the degree (n-1) term.<br>
In this problem, the sum of the roots is therefore {{{(-3a)/1 = -3a}}}.<br>
If the polynomial is a perfect cube, and the sum of the roots is -3a, then each root is -a, and the polynomial is {{{(x+a)^3 = x^3+3ax^2+3a^2x+a^3}}}<br>
So the coefficient of x is {{{b = 3a^2}}}, and the constant is {{{c = a^3}}}.<br>
And then {{{b^3 = (3a^2)^3 = 27a^6 = 27(a^3)^2 = 27c^2}}}