If `a, b, c` are real numbers, then prove that $ \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = -2(a^3+b^3+c^3-3abc) $ and hence show that if the value of this determinant is zero, then either `a+b+c=0` or `a=b=c`.