<s>[INST] Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$ 
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$

My attempt, 
As I know from the original equation, 
$\alpha+\beta+\gamma=-a$  
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$
Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I
m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot. 
By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot.
 [/INST] If we replace $x$ by $x^{1/3}$, we obtain $x + ax^{2/3} + bx^{1/3} + c = 0$. This equation does have the required roots, but it's not a polynomial. 
To get the good ol' cubic form, write it as $x + c = -(ax^{2/3} + bx^{1/3})$ and cube both sides so that$$(x + c)^3 = -(ax^{2/3} + bx^{1/3})^3 = -(a^3 x^2 + b^3 x + 3abx(ax^{2/3} + bx^{1/3})) = -(a^3 x^2 + b^3 x - 3abx(x+c))$$
This method wouldn't generate extraneous roots because we're not squaring anything here.
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