An Eigen Value of $\tiny \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is : Let $a,b,c,d$ be distinct non zero real numbers with $a+b=c+d.$ Then, an eigen value of the matrix $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :
$(i)~a+c ~~(ii)~a+b~~(iii)~a-b~~(iv)~(b-d$
Attempt:
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Solving $\det ( A - \lambda I) = 0$ results in a third degree equation which is difficult to solve. 
Could anyone tell me of an easier method to solve it. Thank you very much for your help in this regard.

Thank you all for your replies. While I was just reflecting, I thought of the following solution, so, thought of sharing the following solution :
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Adding the first two columns and placing it in the first results in no change in the determinant value : 
Hence : 
$\det ( A- \lambda I) = \det  \begin{pmatrix} a+b-\lambda&b&1 \\ c+d-\lambda&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
$= (a+b-\lambda) \det  \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix} =  (c+d-\lambda) \det  \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
Hence, $\det  ( A- \lambda I) =0 \implies a+b = \lambda = c+d$ 
