Use row reductions to show $det(T)=0$

Use row operations to show that $det(T)=0$, where

$$T = \begin{bmatrix} x^2 & 2x+1 & 4x+4 & 6x+9\\ y^2 & 2y+1 & 4y+4 & 6y+9\\ z^2 & 2z+1 & 4z+4 & 6z+9\\ w^2 & 2w+1 & 4w+4 & 6w+9 \end{bmatrix}$$ I am asked to use row operations to show that determinant is equal to 0. I don't see how it is possible because you cant subtract any row with another row as each have different variables. Please someone guide me.

If $c_i$ stands for ith column, just observe that $3\cdot c_3 - 3\cdot c_2 = c_4$. When a matrix has linearly dependent columns or rows, its determinant will be zero.

@phdstudent has shown you how to do it with column operations.

If you have to do it with row operations, replace the $i$-th row with the difference between the $i$-th row and the ($i+1$)-th row, (taking $i + 1 = 1$ when $i = 4$) to get $$ \det(T) = (x - y) \cdot (y - z) \cdot (z - w) \cdot(w - x) \cdot \det \left(\begin{bmatrix} x + y & 2 & 4 & 6\\ y + z & 2 & 4 & 6\\ z + w & 2 & 4 & 6\\ w + x & 2 & 4 & 6 \end{bmatrix}\right). $$ The last three columns of the new matrix are now clearly pairwise dependent. But if you have to do everything by rows, note that in the new matrix, the sum of the even rows equals the sum of the odd ones, so the rows are dependent, and the determinant is zero. Or, if you prefer, do Laplace expansion with respect to the first column.