Using properties of determinants, show that \begin{vmatrix} b+c & a & a \ b & c+a & b \ c & c & a+b \end{vmatrix} = 4abc

Let det(A)=\begin{vmatrix} b+c & a & a \ b & c+a & b \ c & c & a+b \end{vmatrix}=(b+c)[(c+a)(a+b)-bc]-a[b(a+b)-bc]+a[bc-c(c+a)]=(b+c)(ac+a^2+ab+bc-bc)-a(ab+b^2-bc)+a(bc-c^2-ac)=(b+c)(ac+a^2+ab)-a^2b-ab^2+abc+abc-ac^2-a^2c=abc+a^2b+ab^2+ac^2+a^2c+abc-a^2b-ab^2+abc+abc-ac^2-a^2c=4abc