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.

Let $p_k = \alpha^k + \beta^k + \gamma^k$. By Newton identities, we have
$$\begin{align}
p_1 + a = 0 &\implies p_1 = -a\\
p_2 + ap_1 + 2b = 0&\implies p_2 = a^2 - 2b\\
p_3 + ap_2 + bp_1 + 3c = 0&\implies p_3 = -a^3 + 3ab - 3c
\end{align}
$$
In particular, we have
$$\alpha^3 + \beta^3 + \gamma^3 = p_3 =  -a^3 + 3ab - 3c$$
Apply Newton identities to the polynomial
$$x^3 + \frac{b}{c} x^2 + \frac{a}{c} x + \frac{1}{c}$$
which have roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$, we get
$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} = 
-\frac{b^3}{c^3} + 3\frac{ba}{c^2} - \frac{3}{c}\tag{*1}$$
By Vieta's formula, we have $\alpha\beta\gamma = - c$. Multiply LHS of $(*1)$ by $\alpha^3\beta^3\gamma^3$ and RHS by the same number $-c^3$, we get
$$\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 3c^2 - 3abc  + b^3$$
As a result, the polynomial with roots $\alpha^3, \beta^3, \gamma^3$ is
( by Vieta's formula again)
$$x^3 - (\alpha^3+\beta^3+\gamma^3) x^2 + (\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3) - \alpha^3\beta^3\gamma^3\\
= x^3 + (a^3 -3ab + 3c)x^2 + (3c^2 - 3abc  + b^3) x + c^3
$$
How to remember the Newton identifies
In general, given a polynomial of the form
$$P(x) = x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n$$
with roots $\lambda_1, \ldots, \lambda_n$. The sequence of numbers $p_k = \sum_{i=1}^n\lambda_i^n$ satisfies a bunch of identities.
$$\begin{cases}
p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{k-1} p_1 + \color{red}{k} a_k = 0, & k \le n\\
p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{n-1} p_{k-n+1} + a_n p_{k-n} = 0,
& k > n
\end{cases}$$
For any particular $k$, you can obtain the corresponding identity by multiplying $p_{k-\ell}$ term with $a_\ell$ and sum over all available $0 \le \ell \le n$. 
When $k \le n$, you will have terms like $p_0$, $p_{-1}$,... Just replace all appearance of $p_0$ by $\color{red}{k}$ and forget all $p_\ell$ with negative $\ell$.
