Without directly evaluting, show that the determinant of $A = 0$ 
Without directly evaluting, show that 
  $det \left[
       \begin{array}{ccc}
         b + c & c + a & b + a \\
           a & b & c \\
           1 & 1 & 1 
       \end{array}
     \right]
=0$  

I am stuck on this one. I can only do this by evuating.
Things that I know:
1) Square matrix $A$ with two proportional rows or columns as a $det(A) = 0$
2) Square matrix $A$ has $det(A)=0$ if it has a row or column of zeros.
Can anyone help?
Thanks.

Here's one possibility: For a fixed value of $a,b$, $\det(A)$ is a degree two polynomial in $c$. It is obviously $0$ for $c=0,a,b$, and hence the polynomial is identically $0$. (Unless $a=b$, $a=0$, etc, but then you can use the same reasoning with the other variables).
