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.
 A: 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 $f(x)=x^2+2x+1$

A: Let the two roots of $g(x)$ be $p$ and $q$. Since roots of $f(x)$ are cubes of the roots of $g(x)$,
$$\begin{align*}
p+q &= -c\tag1\\
pq &= d\tag2\\
p^3 + q^3 &= -a\tag3\\
p^3q^3 &= b\tag4
\end{align*}$$
From $(3)$, 
$$\begin{align*}
-a &= p^3 + q^3\\
&= (p+q)(p^2 - pq+q^2)\\
&= (p+q)(p^2 + 2pq + q^2 - 3pq)\\
&= -c[(-c)^2 -3d]\\
a &= c^3 -3cd\tag5
\end{align*}$$
And from $(2)$ and $(4)$, $$b = p^3 q^3 = d^3\tag6$$
Since the sum and the product of roots of $g(x)$ are equal, $$d = -c \tag7$$
Writing $a$ and $b$ in terms of $c$,


*

*Substitute $(7)$ into $(5)$, $$a = c^3 + 3c^2\tag8$$

*Substitute $(7)$ into $(6)$, $$b = -c^3\tag{9}$$
Since $f(x)$ has double roots, by using $(8)$ and $(9)$,
$$\begin{align*}
\Delta_f = a^2 - 4b &= 0\\
(c^3 + 3c^2)^2 + 4c^3 &= 0\\
c^4 (c+3)^2 + 4c^3 &= 0\\
c^3 (c^3 + 6c^2 + 9c + 4) &= 0\\
c^3 (c+1)^2(c+ 4) &= 0\\
\end{align*}$$
If $c \in \{0, -4\}$ and using $(7)$, 
$$\begin{align*}
\Delta_g &= c^2 - 4d\\
&= c^2 + 4c\\
&= c(c+4)\\
&= 0
\end{align*}$$ which means $g(x)$ has a double root, contradicting to the given conditions.
Hence $c = -1$. Then $a = 2$, $b=1$ and $d = 1$.
A: 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 $g(x)=(x+\omega)(x+\omega^2)=x^2-x+1$.
