The value of $\begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(c+a) \\ 1 & ab & ab(a+b) \end{vmatrix}$ is

$\begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(c+a) \\ 1 & ab & ab(a+b) \end{vmatrix}$

Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$= \begin{vmatrix} 1 & bc & bc(b+c) \\ 0 & ca - bc & ca(c+a) - bc(b+c) \\ 0 & ab - bc & ab(a+b) - bc(b+c) \end{vmatrix}$

For $R_2$:

$ca - bc = c(a - b)$

$ca(c+a) - bc(b+c)$

$= c(a^2 + ac - b^2 - bc)$

$= c[(a^2 - b^2) + c(a - b)]$

$= c(a-b)(a+b+c)$

For $R_3$:

$ab - bc = b(a - c)$

$ab(a+b) - bc(b+c)$

$= b(a^2 + ab - c^2 - bc)$

$= b[(a^2 - c^2) + b(a - c)]$

$= b(a-c)(a+b+c)$

$= \begin{vmatrix} 1 & bc & bc(b+c) \\ 0 & c(a-b) & c(a-b)(a+b+c) \\ 0 & b(a-c) & b(a-c)(a+b+c) \end{vmatrix}$

Expanding along $C_1$:

$= 1 \times \begin{vmatrix} c(a-b) & c(a-b)(a+b+c) \\ b(a-c) & b(a-c)(a+b+c) \end{vmatrix}$

In this $2 \times 2$ matrix, $C_2 = (a+b+c) \times C_1$.

Since two columns are proportional, the determinant equals $\boxed{0}$.