# 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?

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

• Thank You @Atticus this didnt click me earlier, let me try to get the answer now. Mar 14 '20 at 8:10
• Ok, thanks, I got it now, you decomposed the last column! Thanks you, nice trick! Mar 14 '20 at 8:14
• Got the answer :) Mar 14 '20 at 8:18
• That's ok, it was a really nice solution. Taught an old dog a new trick C: Mar 14 '20 at 8:18
• @Atticus I have provided an alternative answer. Mar 14 '20 at 11:24

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...).

• This is neat (+1).
– LHF
Mar 14 '20 at 11:39
• Thank You for providing a different approach. Nice Method :) Mar 14 '20 at 11:42

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.

• Thank You! Thats smart and super helpful :D Mar 14 '20 at 16:34