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Let $P,Q,R$ be polynomials of degree less than 2 and $a,b,c \in \mathbb{C}$

such that $D(a,b,c)=\begin{vmatrix}P(a)&&Q(a)&&R(a)\\P(b)&&Q(b)&&R(b)\\P(c)&&Q(c)&&R(c)\end{vmatrix}=1$

Prove that $D(1,b,c)+D(a,1,c)+D(a,b,1)=D(a,b,c)$.

My approach: Consider $P(x)=mx^2+nx+p, Q(x)=dx^2+ex+f, R(x)=rx^2+sx+t$ Then we have $D(a,b,c)=-\begin{vmatrix}m&&n&&p\\ d&&e&&f\\r&&s&&t\end{vmatrix}\dot{}\begin{vmatrix}1&&1&&1\\ a&&b&&c\\a^2&&b^2&&c^2\end{vmatrix}$

Then we need to show that $\begin{vmatrix}1&&1&&1\\ a&&b&&c\\a^2&&b^2&&c^2\end{vmatrix}=\begin{vmatrix}1&&1&&1\\ 1&&b&&c\\1&&b^2&&c^2\end{vmatrix}+\begin{vmatrix}1&&1&&1\\ a&&1&&c\\a^2&&1&&c^2\end{vmatrix}+ \begin{vmatrix}1&&1&&1\\ a&&b&&1\\a^2&&b^2&&1\end{vmatrix}$

I did this by considering the system of equations

$x+y+z=1 \\ ax+by+cz=1 \\a^2x+b^2y+c^2z=1$

and using Cramer rule. I am interested if there is another approach.

StubbornAtom
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amarius8312
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    You wrote that the degree of $P, Q, R$ is less than two, but then assumed them to be quadratic in your approach. Which is it? – Tom Apr 12 '19 at 16:16
  • I didn;t assume this, the leading coefficients can be zero so the degree can be lower – amarius8312 Apr 12 '19 at 19:47

1 Answers1

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Consider $$D(x,b,c)+D(a,x,c)+D(a,b,x)-D(a,b,c)$$ It is a quadratic polynomial in $x$, put $x=a$. The expression becomes, $$D(a,b,c)+D(a,a,c)+D(a,b,a)-D(a,b,c)$$ The second and third terms are zero due to identical rows, and first and third cancel. So $x=a$ is a root of this quadratic polynomial. Similarly, $x=b$ and $x=c$ are also roots of this. Since, it is a quadratic polynomial having three roots, it is identically zero for all $x\in \mathbb R$, hence holds for $x=1$.

Hope it helps:)

Martund
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