Finding xy+yz+zx such that the given determinant = 0

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?

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

A solution in a different spirit :

We are going to assume that none of the values $x,y$ or $z$ is zero (otherwise, the expression (D) would be meaningless).

Let

$$M=\begin{pmatrix}x^4-1&x^3&x\\y^4-1&y^3&y\\z^4-1&z^3&z\end{pmatrix}$$

(I have modified the order of columns, WLOG).

As $\det(M)=0$, there exists a linear dependency on the columns of $M$, i.e., there exists $a,b,c$ such that

$$\begin{cases}a(x^4-1)+bx^3+cx&=&0\\a(y^4-1)+by^3+cy&=&0\\a(z^4-1)+bz^3+cz&=&0\end{cases}$$

Let us assume $a \neq 0$ (see remark below). WLOG, we can assume that $a=1$.

As a consequence, 4th degree polynomial :

$$P(t):=t^4+bt^3+0t^2+ct-1$$

has $x,y,z$ for its roots ; let us denote by $r$ the fourth root.

Now, let us write the second and the last Viète formulas (those which do not involve an unknown letter) :

$$\left\{\begin{array}{rcr}xy+yz+zx+rx+ry+rz&=&0\\xyzr&=&-1\end{array}\right.$$

$$r=- \dfrac{xy+yz+zx}{x+y+z}=- \dfrac{1}{xyz}$$

therefore proving result D.

Remark : In fact $a$ cannot be $0$. Otherwise, it would mean that there is a linear dependency between the two first columns. Should column 2 be a multiple of column 1, their resp. entries would be proportional, i.e.,

$$\dfrac{x^3}{x}=\dfrac{y^3}{y}=\dfrac{z^3}{z}$$

$$\iff \ \ x^2=y^2=z^2 $$

which is not possible for different $x,y,z$ (2 at least would be equal...).