<s>[INST] Find coefficients of polynomials $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$ $(a,b,c,d \in \mathbb{R})$ Roots of polynomial $f(x)=x^2+ax+b$ are cubes of the roots of polynomial $g(x)=x^2+cx+d$. Sum and product of roots of polynomial $g(x)$ are equal. Find coefficients $a,b,c,d$ so that polynomial $f(x)$ has double root, and $g(x)$ doesn't have double root.
From the condition that $f(x)$ has double root, discriminant of quadratic equation $x^2+ax+b=0$ must be zero, $$D=a^2-4b=0,b=\frac{a^2}{4}$$ 
Sum and product of roots of polynomial $g(x)$ are equal, $$x_1^3+x_2^3=x_1^3x_2^3$$
Vieta's formulas for two polynomials,
$$x_1+x_2=-a$$
$$x_1x_2=b$$
$$x_1^3+x_2^3=-c$$
$$x_1^3x_2^3=d$$
If we cube first equation, $$(x_1^3+x_2^3)+3x_1x_2(x_1+x_2)=-a^3$$
$$-c-3ba=-a^3$$
From $b=\frac{a^2}{4}$ $$a^3-4c=0,c=\frac{a^3}{4}$$
From $x_1^3+x_2^3=x_1^3x_2^3$ implies $d=-c$
If we cube second equation, $$x_1^3x_2^3=b^3=\frac{a^6}{64}=d=-c$$
I don't know how to find these coefficients.
Thanks for replies.
 [/INST] The roots of $f$ are equal, and the coefficients of $f$ are real. So we must have $f(x)=(x-\alpha)^2$ for some real $\alpha$.
The roots of $g$ are all cube roots of $\alpha$, so they must be drawn from the set $\{\sqrt[3]{\alpha},\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}\}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is a primitive cube root of unity. Note that $\sqrt[3]{\alpha}$ is the only real number in this set; since the roots of $g$ are distinct, it follows that they cannot all be real. Since $g$ has real coefficients, they must be complex conjugates; thus $g$ has roots $\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}$.
Now, as the sum and product of the roots of $g$ is equal, we have
$$(\omega+\omega^2)\sqrt[3]{\alpha}=\omega^3\sqrt[3]{\alpha^2}\\
-\sqrt[3]{\alpha}=\sqrt[3]{\alpha^2}$$
and since $\alpha$ is real this implies that $\alpha^2=-\alpha$, and so $\alpha=-1$. That is, $f(x)=(x+1)^2=x^2+2x+1$, and $g(x)=(x+\omega)(x+\omega^2)=x^2-x+1$.
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