Prove that the determinant of the matrix [[b+c, a, a], [b, c+a, b], [c, c, a+b]] is 2abc.

Let the given determinant be $\Delta$.

$$\Delta = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$$

This operation changes the first row while keeping the determinant value unchanged.
The new elements of the first row will be:

• First element: $(b+c) - b - c = 0$
• Second element: $a - (c+a) - c = a - c - a - c = -2c$
• Third element: $a - b - (a+b) = a - b - a - b = -2b$

$$\Delta = \begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$$

The determinant of a $3 \times 3$ matrix $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ is $a(ei-fh) - b(di-fg) + c(dh-eg)$.

$$\Delta = 0 \cdot \begin{vmatrix} c+a & b \\ c & a+b \end{vmatrix} - (-2c) \cdot \begin{vmatrix} b & b \\ c & a+b \end{vmatrix} + (-2b) \cdot \begin{vmatrix} b & c+a \\ c & c \end{vmatrix}$$ $$\Delta = 0 + 2c \cdot (b(a+b) - b \cdot c) - 2b \cdot (b \cdot c - c(c+a))$$

$$\Delta = 2c \cdot (ab + b^2 - bc) - 2b \cdot (bc - c^2 - ac)$$ $$\Delta = 2abc + 2b^2c - 2bc^2 - (2b^2c - 2bc^2 - 2abc)$$ $$\Delta = 2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc$$

$$\Delta = (2abc + 2abc) + (2b^2c - 2b^2c) + (-2bc^2 + 2bc^2)$$ $$\Delta = 4abc + 0 + 0$$ $$\Delta = 4abc$$

This matches the right-hand side (RHS) of the given equation.

Therefore, it is proven that:

$$\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = \boxed{4abc}$$