Finding xy+yz+zx such that the given determinant = 0 
$x≠y≠z$
$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
  Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ | 

Given Ans -  D
What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0.....but this way the opened determinant is still too complex
What I did second was putting values of x and y but with that I was only able to eliminate option A & B
I need help with the correct approach (the correct row transformation) or any other method I can try. 

Hint:
$$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}=\begin{vmatrix}x&x^3&x^4\\y&y^3&y^4\\z&z^3&z^4\end{vmatrix}-\begin{vmatrix}x&x^3&1\\y&y^3&1\\z&z^3&1\end{vmatrix}=xyz\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}-\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}$$
Can you end it from here?

 $$\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$$
$$\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(x+y+z)$$

