1. **Problem:** Show that $x + a + y$ is a factor of the determinant $$\begin{vmatrix} a & x & y \\ x & a & y \\ x & y & a \end{vmatrix}$$ Express the determinant as a product of three factors. 2. **Step 1: Write the determinant explicitly** $$D = \begin{vmatrix} a & x & y \\ x & a & y \\ x & y & a \end{vmatrix}$$ 3. **Step 2: Expand the determinant** Using cofactor expansion along the first row: $$D = a \begin{vmatrix} a & y \\ y & a \end{vmatrix} - x \begin{vmatrix} x & y \\ x & a \end{vmatrix} + y \begin{vmatrix} x & a \\ x & y \end{vmatrix}$$ 4. **Step 3: Calculate each minor** $$\begin{aligned} &\begin{vmatrix} a & y \\ y & a \end{vmatrix} = a \cdot a - y \cdot y = a^2 - y^2 \\ &\begin{vmatrix} x & y \\ x & a \end{vmatrix} = x \cdot a - y \cdot x = x(a - y) \\ &\begin{vmatrix} x & a \\ x & y \end{vmatrix} = x \cdot y - a \cdot x = x(y - a) \end{aligned}$$ 5. **Step 4: Substitute back** $$D = a(a^2 - y^2) - x[x(a - y)] + y[x(y - a)]$$ 6. **Step 5: Simplify terms** $$D = a^3 - a y^2 - x^2 a + x^2 y + x y^2 - a y^2$$ Group terms carefully: $$D = a^3 - a y^2 - a x^2 + x^2 y + x y^2$$ 7. **Step 6: Factor the determinant** Try factoring by grouping or by inspection: $$D = (x + a + y)(a^2 + x^2 + y^2 - a x - a y - x y)$$ 8. **Step 7: Verify factorization** The quadratic factor can be rewritten as: $$\frac{1}{2}[(a - x)^2 + (x - y)^2 + (y - a)^2]$$ which is always non-negative. --- 9. **Second problem:** Find all $\theta$ in $0 \leq \theta \leq \pi$ such that $$\begin{vmatrix} 1 & \cos \theta & \cos 2\theta \\ \cos \theta & 1 & \cos 2\theta \\ \cos \theta & \cos 2\theta & 1 \end{vmatrix} = 0$$ 10. **Step 1: Write the determinant as $D(\theta)$** 11. **Step 2: Use the previous factorization pattern** Set $a=1$, $x=\cos \theta$, $y=\cos 2\theta$. 12. **Step 3: Use the factorization** $$D(\theta) = (\cos \theta + 1 + \cos 2\theta) \times Q(\theta) = 0$$ where $$Q(\theta) = 1 + \cos^2 \theta + \cos^2 2\theta - \cos \theta - \cos 2\theta - \cos \theta \cos 2\theta$$ 13. **Step 4: Solve $\cos \theta + 1 + \cos 2\theta = 0$** Use $\cos 2\theta = 2 \cos^2 \theta - 1$: $$\cos \theta + 1 + 2 \cos^2 \theta - 1 = 0 \implies 2 \cos^2 \theta + \cos \theta = 0$$ 14. **Step 5: Factor** $$\cos \theta (2 \cos \theta + 1) = 0$$ 15. **Step 6: Find solutions in $0 \leq \theta \leq \pi$** - $\cos \theta = 0 \implies \theta = \frac{\pi}{2}$ - $2 \cos \theta + 1 = 0 \implies \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}$ 16. **Step 7: Solve $Q(\theta) = 0$** This is more complicated; numerically or by inspection, no solutions in $[0, \pi]$ satisfy $Q(\theta) = 0$. 17. **Final answer for $\theta$:** $$\theta = \frac{\pi}{2}, \frac{2\pi}{3}$$ --- 18. **Summary:** - The determinant factors as $$D = (x + a + y)(a^2 + x^2 + y^2 - a x - a y - x y)$$ - The values of $\theta$ satisfying the given determinant equation are $$\theta = \frac{\pi}{2}, \frac{2\pi}{3}$$