Let the given quadratic polynomial $p\left(x\right)=x^2+ax+b$ has two distinct real roots $\alpha$ and $\beta$, then $p\left(x\right)=x^2+ax+b=\left(x-\alpha \right)\left(x-\beta \right)$

and since $g\left(x\right)=p\left(x^3\right)=\left(x^3-\alpha \right)\left(x^3-\beta \right)$

let $\alpha ={\alpha }_1^3$ and $\beta ={\beta }_1^3$

then $g\left(x\right)=\left(x^3-{\alpha }_1^3\right)\left(x^3-{\beta }_1^3\right)$

$=\left(x-{\alpha }_1\right)\left(x-{\beta }_1\right)\left(x^2+{\alpha }_{1}x+{\alpha }_1^2\right)\left(x^2+{\beta }_{1}x+{\beta }_1^2\right)$

$\because$  the discriminants of quadratic equations $x^2+{\alpha }_{1}x+{\alpha }_1^2$ and $x^2+{\beta }_{1}x+{\beta }_1^2$ are

negative.

$\therefore g\left(x\right)$ has exactly two distinct real roots and since $g\left(x\right)=x^6+a{x}^3+b$ is an even degree polynomial, so there exists a real number $\alpha$ ' such that $g\left(x\right)\ge \alpha$ for all real $x$