<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] Hint: Let $\alpha, \beta$ be the roots of $g(x)=0$ then the roots of $f(x)$ are $\alpha^3=\beta^3$ which means that $\alpha^3-\beta^3=0=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)$ and since $\alpha\neq \beta$ you have $\alpha^2+\alpha\beta+\beta^2=0$
Then plug in the condition $\alpha \beta=\alpha+\beta=r$ so that $\alpha^2+\alpha\beta+\beta^2=r^2-r=0$

Since others have completed the solution, here's my take on a simpler answer using the hint.

 You can't have $r=0$, because this would give a double root of $0$, so it must be that $r=1$. Then $g(x)=x^2-x+1$. It is clear that $b=\alpha^3\beta^3=r^3=1$, and then $-a=\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)=r(r^2-3r)=-2$ so that $a=2$ so that $f(x)=x^2+2x+1$

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