Without expanding the determinant prove that

|1 ab c(a+b)|
|1 bc a(b+c)|
|1 ca b(c+a)|

=0

Add the second two columns to get
ab + ac + bc
ab + ac + bc
ab + ac + bc

which is obviously linearly dependent with the first column. The columns are not linearly independent, so the matrix is not invertible, so its determinant is 0.