1. **Problem Statement:** We want to verify the determinant equality: $$\begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} = 4a^2b^2c^2$$ 2. **Matrix and Goal:** The matrix is: $$M = \begin{bmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{bmatrix}$$ We want to find $\det(M)$ and show it equals $4a^2b^2c^2$. 3. **Step 1: Factor out $abc$ from rows or columns** Notice each element contains factors of $a$, $b$, or $c$. We factor $abc$ from the determinant: $$\det(M) = abc \times \det\begin{bmatrix} -a & a & a \\ b & -b & c \\ c & c & -c \end{bmatrix}$$ 4. **Step 2: Perform column operations to simplify** Apply column operations: $$c'_1 = c_1 + c_2$$ $$c'_2 = c_2 + c_3$$ This yields: $$\det = abc \times \det\begin{bmatrix} 0 & 2a & a \\ 0 & 0 & b \\ 2c & 0 & -c \end{bmatrix}$$ 5. **Step 3: Calculate the determinant of the simplified matrix** Calculate: $$\det = abc \times \left(0 \times \begin{vmatrix} 0 & b \\ 0 & -c \end{vmatrix} - 2a \times \begin{vmatrix} 0 & b \\ 2c & -c \end{vmatrix} + a \times \begin{vmatrix} 0 & 0 \\ 2c & 0 \end{vmatrix}\right)$$ 6. **Step 4: Evaluate minors** - First minor: $0$ (since first element is zero) - Second minor: $$\begin{vmatrix} 0 & b \\ 2c & -c \end{vmatrix} = 0 \times (-c) - b \times 2c = -2bc$$ - Third minor: $$\begin{vmatrix} 0 & 0 \\ 2c & 0 \end{vmatrix} = 0$$ 7. **Step 5: Substitute minors back** $$\det = abc \times (0 - 2a \times (-2bc) + a \times 0) = abc \times (4abc) = 4a^2b^2c^2$$ 8. **Conclusion:** We have shown step-by-step that: $$\det(M) = 4a^2b^2c^2$$ which matches the right-hand side (RHS) of the original equation. This completes the proof.