The roots of $x^3 + a x^2 + b x + c = 0$ are $\alpha, \beta$ and $\gamma.$ Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3.$
1. **Using Vieta's Formulas:**
   Given the polynomial \(x^3 + a x^2 + b x + c = 0\) with roots \(\alpha, \beta, \gamma\), Vieta's formulas give us:
   \[
   \alpha + \beta + \gamma = -a
   \]
   \[
   \alpha \beta + \beta \gamma + \gamma \alpha = b
   \]
   \[
   \alpha \beta \gamma = -c
   \]

2. **Form of the Desired Polynomial:**
   Let the desired polynomial be \(x^3 - dx^2 + ex - f\), where:
   \[
   d = \alpha^3 + \beta^3 + \gamma^3
   \]
   \[
   e = \alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3
   \]
   \[
   f = \alpha^3 \beta^3 \gamma^3
   \]

3. **Finding \(f\):**
   \[
   f = (\alpha \beta \gamma)^3 = (-c)^3 = -c^3
   \]

4. **Finding \(d\):**
   Using the identity for the sum of cubes:
   \[
   \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha) + 3 \alpha \beta \gamma
   \]
   We first find \(\alpha^2 + \beta^2 + \gamma^2\):
   \[
   (\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha \beta + \beta \gamma + \gamma \alpha)
   \]
   \[
   (-a)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2b
   \]
   \[
   a^2 = \alpha^2 + \beta^2 + \gamma^2 + 2b
   \]
   \[
   \alpha^2 + \beta^2 + \gamma^2 = a^2 - 2b
   \]
   Now substitute back:
   \[
   \alpha^3 + \beta^3 + \gamma^3 = (-a)(a^2 - 2b - b) + 3(-c)
   \]
   \[
   \alpha^3 + \beta^3 + \gamma^3 = -a(a^2 - 3b) - 3c
   \]
   \[
   \alpha^3 + \beta^3 + \gamma^3 = -a^3 + 3ab - 3c
   \]
   Thus, \(d = -a^3 + 3ab - 3c\).

5. **Finding \(e\):**
   Using the identity for the product of squares:
   \[
   (\alpha \beta + \beta \gamma + \gamma \alpha)^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 + 2\alpha \beta \gamma (\alpha + \beta + \gamma)
   \]
   \[
   b^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 + 2(-c)(-a)
   \]
   \[
   b^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 + 2ac
   \]
   \[
   \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 = b^2 - 2ac
   \]
   Now, we need to find \(\alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3\):
   \[
   (\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2)(\alpha \beta + \beta \gamma + \gamma \alpha) = \alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3 + \alpha \beta \gamma (\alpha^2 \beta + \alpha^2 \gamma + \beta^2 \alpha + \beta^2 \gamma + \gamma^2 \alpha + \gamma^2 \beta)
   \]
   Using the previous result for \(\alpha^2 \beta + \alpha^2 \gamma + \beta^2 \alpha + \beta^2 \gamma + \gamma^2 \alpha + \gamma^2 \beta\):
   \[
   \alpha^2 \beta + \alpha^2 \gamma + \beta^2 \alpha + \beta^2 \gamma + \gamma^2 \alpha + \gamma^2 \beta = -a^3 + 2ab - (\alpha^3 + \beta^3 + \gamma^3) = 3c - ab
   \]
   Substituting back:
   \[
   \alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3 + (-c)(3c - ab) = b^3 - 2abc
   \]
   \[
   \alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3 - 3c^2 + abc = b^3 - 2abc
   \]
   \[
   \alpha^3 \beta^3 + \beta^3 \gamma^3 + \gamma^3 \alpha^3 = b^3 + 3c^2 - 3abc
   \]
   Thus, \(e = b^3 + 3c^2 - 3abc\).

6. **Forming the Polynomial:**
   The desired polynomial is:
   \[
   x^3 - dx^2 + ex - f
   \]
   Substituting the values of \(d\), \(e\), and \(f\):
   \[
   x^3 + (a^3 - 3ab + 3c)x^2 + (b^3 + 3c^2 - 3abc)x + c^3
   \]

The final answer is \(\boxed{x^3 + (a^3 - 3ab + 3c)x^2 + (b^3 + 3c^2 - 3abc)x + c^3}\)