Suppose that the roots of the quadratic $x^2 + ax + b$ are $\alpha$ and $\beta$. Then $\alpha^3$ and  $\beta^3$ are the roots of some quadratic $x^2 + cx + d$. Find $c$ in terms of $a$ and $b$.

1. By Vieta's formulas for the quadratic equation \(x^2 + ax + b = 0\), the sum and product of the roots \(\alpha\) and \(\beta\) are given by:
   \[
   \alpha + \beta = -a
   \]
   \[
   \alpha \beta = b
   \]

2. We need to find the quadratic equation whose roots are \(\alpha^3\) and \(\beta^3\). Let this quadratic equation be \(x^2 + cx + d = 0\).

3. By Vieta's formulas for the new quadratic equation, the sum and product of the roots \(\alpha^3\) and \(\beta^3\) are given by:
   \[
   \alpha^3 + \beta^3 = -c
   \]
   \[
   \alpha^3 \beta^3 = d
   \]

4. To find \(\alpha^3 + \beta^3\) in terms of \(a\) and \(b\), we use the identity for the sum of cubes:
   \[
   \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)
   \]

5. First, we express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha \beta\):
   \[
   \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta
   \]
   Substituting \(\alpha + \beta = -a\) and \(\alpha \beta = b\):
   \[
   \alpha^2 + \beta^2 = (-a)^2 - 2b = a^2 - 2b
   \]

6. Now, substitute \(\alpha + \beta\) and \(\alpha^2 + \beta^2\) into the sum of cubes identity:
   \[
   \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta)
   \]
   \[
   \alpha^3 + \beta^3 = (-a)((a^2 - 2b) - b)
   \]
   \[
   \alpha^3 + \beta^3 = (-a)(a^2 - 3b)
   \]
   \[
   \alpha^3 + \beta^3 = -a^3 + 3ab
   \]

7. Therefore, we have:
   \[
   -c = -a^3 + 3ab
   \]
   \[
   c = a^3 - 3ab
   \]

The final answer is \(\boxed{a^3 - 3ab}\).