Mathematical Enigma - XI

The first statement is equivalent to saying that $a$ (multiplicity 2) is a root of the equation $x^3+3px+q=0$ .

By Vieta's formulas, the sum of the three roots of the equation is 0. Letting $x_3$ to be the third root, we have $x_1+x_2+x_3=a+a+x_3=0\implies x_3=-2a$

Furthermore, the product of the three roots is $-q$ , and the sum of two-wise products of the roots is $3p$ . So $x_1 x_2 x_3=-q\implies q=-(a)(a)(-2a)=2a^3$ $x_1 x_2 + x_1 x_3 + x_2 x_3 = 3p \implies p = \frac{1}{3}[(a)(a)+(a)(-2a)+(a)(-2a)]=-a^2$ Therefore, $q^2 + 4p^3 + 7 = (2a^3)^2 + 4(-a^2)^3 + 7 = 4a^6 - 4a^6 + 7 = \boxed{7}$

let $f(x) = x^3 + 3px + q$

we have $f(a) = 0$ and $f'(a) = 0$

then : $p = -a^2$ and $q = 2a^3$

so that : $q^2 + 4p^3 + 7 = 7$

just put
a=0,p=0 and q=0. every thing satisfied.