The assumption $a^3 + pa + q = 0$ implies $q = -a(a^2 + p)$, whence $x^3 + px + q = (x-a)(x^2 + ax + a^2 + p)$. The discriminant of $x^2 + ax + a^2 + p$ is $D = a^2 - 4(a^2 + p) = -(3a^2 + 4p)$; the assumption $a^2 \le -\frac{4}{3}p$ implies $D \ge 0$. Hence there are real numbers $b$ and $c$ such that $x^2 + ax + a^2 + p = (x-b)(x-c)$, so the polynomial $x^3 + px + q$ has roots $b$ and $c$. If $a = b = c$ then $x^3 + px + q = (x-a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$. Hence $-3a = 0$, $3a^2 = p$ and $a^3 = q$, implying $p = q = 0$. This contradicts the assumption. Consequently, $x^3 + px + q$ has a real root different from $a$.