1. **Stating the problem:** We need to show that the determinant of the matrix $$\begin{vmatrix}-a^2 & 2ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2\end{vmatrix} = 4a^2b^2c^2.$$ 2. **Recall the formula for the determinant of a 3x3 matrix:** For matrix $$\begin{bmatrix}x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33}\end{bmatrix},$$ the determinant is $$x_{11}(x_{22}x_{33} - x_{23}x_{32}) - x_{12}(x_{21}x_{33} - x_{23}x_{31}) + x_{13}(x_{21}x_{32} - x_{22}x_{31}).$$ 3. **Apply the formula to our matrix:** Let $$M = \begin{bmatrix}-a^2 & 2ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2\end{bmatrix}.$$ Calculate each part: - First term: $$-a^2((-b^2)(-c^2) - (bc)(cb)) = -a^2(b^2c^2 - b c^2 b) = -a^2(b^2 c^2 - b^2 c^2) = -a^2 \times 0 = 0.$$ - Second term: $$-2ab(ba(-c^2) - bc(ca)) = -2ab(-a b c^2 - b c a c) = -2ab(-a b c^2 - a b c^2) = -2ab(-2 a b c^2) = 4 a^2 b^2 c^2.$$ - Third term: $$ac(ba(cb) - (-b^2)(ca)) = ac(a b^2 c - (-b^2) a c) = ac(a b^2 c + a b^2 c) = ac(2 a b^2 c) = 2 a^2 b^2 c^2.$$ 4. **Sum all terms:** $$0 + 4 a^2 b^2 c^2 + 2 a^2 b^2 c^2 = 6 a^2 b^2 c^2.$$ 5. **Re-examine the second and third terms carefully:** There was a sign error in the second term calculation. Let's correct it: Second term: $$-x_{12}(x_{21}x_{33} - x_{23}x_{31}) = -2ab(ba(-c^2) - bc(ca)) = -2ab(a b (-c^2) - b c a c) = -2ab(-a b c^2 - a b c^2) = -2ab(-2 a b c^2) = 4 a^2 b^2 c^2.$$ Third term: $$x_{13}(x_{21}x_{32} - x_{22}x_{31}) = ac(ba(cb) - (-b^2)(ca)) = ac(a b^2 c - (-b^2) a c) = ac(a b^2 c + a b^2 c) = 2 a^2 b^2 c^2.$$ 6. **Sum all terms again:** $$0 + 4 a^2 b^2 c^2 + 2 a^2 b^2 c^2 = 6 a^2 b^2 c^2.$$ 7. **Conclusion:** The determinant equals $$6 a^2 b^2 c^2,$$ not $$4 a^2 b^2 c^2$$ as stated. Please verify the problem statement or matrix entries for any typos.