Evaluate the determinant $\begin{vmatrix} y+z&z&y\\z&z+x&x\\y&x&x+y\end {vmatrix}$ Performing the operation $R_1\rightarrow R_1-R_2-R_3$
$$\begin{vmatrix} 0&-2x&-2x \\ y&z+x&x \\ z & x&x+y \end{vmatrix}$$
Pulling $-2x$ out and performing $C_2\rightarrow C_2-C_3$
$$-2x\begin{vmatrix} 0&0&1\\ y&z&x \\ z&-y&x+y \end{vmatrix}$$
$$=-2x(-y^2-z^2)$$
However, this is not given in the options, which are 
$4xyz$, $xyz$, $xy$, $x^3+y^3$
What am I doing wrong?
 A: There is an error in your solution. The third row is not the same in the first step. The first step should be
$$\begin{vmatrix} 0&-2x&-2x \\ z&z+x&x \\ y & x&x+y \end{vmatrix}$$
A: You made an error right off the bat: $y$ and $z$ in the first column have somehow gotten swapped after your first manipulation.  
However, instead of manipulating the matrix as is, it’s often fruitful and less work to examine special cases first. Set $x=0$ and observe that the rows of the resulting matrix are linearly dependent. The same is true of the other two variables, so the determinant must have the form $kxyz$ for some unknown constant $k$. You can determine the constant by evaluating the determinant for some other convenient combination of values, such as $x=y=z=1$, for which you can use the well-known eigenvalues of a matrix of the form $aI+b\mathbf 1$ to compute it without expanding further. Since this is multiple-choice, however, it’s pretty obvious that the determinant of $I+\mathbf 1$ isn’t $1$, so the only remaining possibility is $4xyz$.
