<s>[INST] 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. 
 [/INST] We can take advantage of this being a multiple choice question. 
Expanding out the determinant $$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}$$ will give products that look like $xy^3(z^4-1)$, which give us six degree-$8$ terms (such as $xy^3z^4$) and six degree-$4$ terms (such as $-xy^3$).
We know the determinant is divisible by $(x-y)(y-z)(x-z)$, because if any two of $x,y,z$ are equal, the determinant is $0$. Factoring that out, we should get the difference of a degree-$5$ polynomial and a degree-$1$ polynomial.
The four answers are predicting that the polynomial we have left is a multiple of:
\begin{array}{cc}
    (A) & xy + yz + zx - x - y - z \\
    (B) & xy + yz + zx - xyz \\
    (C) & (xy + yz + zx)(x + y + z) - xyz \\
    (D) & (xy + yz + zx)(xyz) - x - y - z
\end{array}
Only option (D) is the difference of a degree-$5$ polynomial and a degree-$1$ polynomial, so it is the only possibility.
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