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Prove that $$ \begin{vmatrix} 1 & a^2 + bc & a^3 \\ 1 & b^2 + ac & b^3 \\ 1 & c^2 + ab & c^3 \\ \end{vmatrix} =(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$

My work

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I am getting - sign extra , please tell me why I am getting this

Rodrigo de Azevedo
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Aakash Kumar
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2 Answers2

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It is faster to exploit Gaussian elimination:

$$\begin{eqnarray*} \det\begin{pmatrix}1 & a^2+bc & a^3 \\ 1 &b^2 +ac & b^3 \\ 1 & c^2+ab & c^3\end{pmatrix}&=&\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &(b-a)(a+b-c) & (b-a)(b^2+ba+a^2) \\ 0 & (c-a)(a+c-b) & (c-a)(c^2+ca+a^2)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &a+b-c & b^2+ba+a^2 \\ 0 & a+c-b & c^2+ca+a^2\end{pmatrix} \\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ a+c-b & c^2+ca+a^2\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2(c-b) & (c-b)(a+b+c)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)(c-b)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2 & a+b+c\end{pmatrix}\end{eqnarray*}$$

and: $$ (a+b-c)(a+b+c)-2(a^2+ab+b^2) = (a+b)^2-c^2-2(a^2+ab+b^2)=-(a^2+b^2+c^2).$$

An even faster approach is to notice that the determinant is clearly zero if $a=b,a=c$ or $b=c$. So the determinant is $(a-b)(b-c)(c-a)$ times a symmetric second-degree polynomial, that is not hard to find.

Jack D'Aurizio
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    While I agree that the zeros make clear that the determinant is of the purported form, it's not at all obvious to me how one deduces the required second-degree polynomial. Care to elaborate? – Semiclassical Jun 27 '16 at 17:59
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    @Semiclassical: one may study the derivative of the determinant with respect to $a$, $b$ or $c$ (https://en.wikipedia.org/wiki/Jacobi%27s_formula) and apply Euler's formula (http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html) – Jack D'Aurizio Jun 27 '16 at 18:09
  • Terribly sorry for the off-topic, but I don't know where to ask this. Why did you remove your answer from this question. I was in a process of understanding it – Glinka Jun 28 '16 at 11:58
  • @Glinka: my previous answer was simply wrong: that decomposition does not exist by Lebesgue differentiation theorem. – Jack D'Aurizio Jun 28 '16 at 12:04
  • You gave two different answers, the first one was based on parity of the first decimal digit. Both was wrong? – Glinka Jun 29 '16 at 10:17
  • @Glinka: I never mentioned the parity of the first decimal digit, also because that obviously lack uniformity: every real number in the interval $(0.3,0.4)$ is associated with the same parity. I tried to work with continued fractions, but that fails too, due to Baire's theorem, for instance. – Jack D'Aurizio Jun 29 '16 at 13:18
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Alternatively, it is easily seen that the determinant of the given matrix is a homogeneous polynomial $P(a,b,c)$ in $a$, $b$, and $c$ of degree $5$. Furthermore, $P(a,b,c)=0$ if any two of the three inputs are equal. Therefore, $P(a,b,c)$ is divisible by $(b-c)(c-a)(a-b)$. Hence, $$P(a,b,c)=(b-c)(c-a)(a-b)\,Q(a,b,c)$$ for some homogeneous quadratic polynomial $Q(a,b,c)$. It is evident that $Q(a,b,c)$ is symmetric in $a,b,c$, whence $Q(a,b,c)=p\,\left(a^2+b^2+c^2\right)+q\,\left(bc+ca+ab\right)$ for some constants $p,q$. As $$-4=P(-1,0,1)=(-1)(2)(-1)\,(2p-q)$$ and $$-30=P(-1,0,2)=(-2)(3)(-1)\,(5p-2q)\,,$$ we get $p=-1$ and $q=0$. That is, $P(a,b,c)=-(b-c)(c-a)(a-b)\left(a^2+b^2+c^2\right)$.

Edit: Just noticed that Jack D'Aurizio had mentioned this polynomial technique.

Batominovski
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