1. **State the problem:** We are given that $x^2 - px + q$ is a factor of the cubic polynomial $x^3 + 3px^2 + 3qx + r$. We need to show that $q = -2p^2$. 2. **Use the factorization property:** If $x^2 - px + q$ is a factor of $x^3 + 3px^2 + 3qx + r$, then there exists a linear polynomial $x + a$ such that: $$x^3 + 3px^2 + 3qx + r = (x^2 - px + q)(x + a)$$ 3. **Expand the right-hand side:** $$ (x^2 - px + q)(x + a) = x^3 + a x^2 - p x^2 - a p x + q x + a q $$ Simplify by grouping like terms: $$= x^3 + (a - p) x^2 + (q - a p) x + a q$$ 4. **Match coefficients with the left-hand side:** From the original polynomial $x^3 + 3 p x^2 + 3 q x + r$, equate coefficients: - Coefficient of $x^3$: $1 = 1$ - Coefficient of $x^2$: $3 p = a - p$ - Coefficient of $x$: $3 q = q - a p$ - Constant term: $r = a q$ 5. **Solve for $a$ from the $x^2$ term:** $$a - p = 3 p \implies a = 4 p$$ 6. **Use $a$ in the $x$ term equation:** $$3 q = q - a p = q - (4 p) p = q - 4 p^2$$ Rearranged: $$3 q - q = -4 p^2 \implies 2 q = -4 p^2 \implies q = -2 p^2$$ 7. **Conclusion:** We have shown that if $x^2 - p x + q$ is a factor of $x^3 + 3 p x^2 + 3 q x + r$, then $q = -2 p^2$ as required.