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\begin{align}\begin{vmatrix}(b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2\end{vmatrix} = 2abc(a+b+c)^3\end{align}

Determinant proof question- Does anyone have an easy solution to this question?

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closed as off-topic by Grigory M, TZakrevskiy, M Turgeon, Davide Giraudo, Michael Tong Apr 27 '14 at 18:53

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We are not here to do computations for you. Show your work and where you are stuck. In particular, do you not know how to compute a determinant? – Chris Gerig Apr 27 '14 at 17:21
Yes,knew,but when i apply it here it becomes very lengthy,and it cannot be done. – user146181 Apr 27 '14 at 17:35
You can apply the very definition of the determinant. The calcultaions will not be short, but it will be relatively easy. If you want a more elegant way, I'll give you the first step: try subtracting second and third lines from the first one. – TZakrevskiy Apr 27 '14 at 18:28

Here is a slightly more "heuristic" way of getting the answer, which is still correct.

Note that if you put $a=0$ in the determinant, you get:

$$\begin{vmatrix} (b+c)^2 & 0 & 0\\ b^2&c^2&b^2\\c^2&c^2&b^2 \end{vmatrix} = 0$$

As the 2nd and 3rd columns are multiples of each other. So you know that $a$ must be a root of the expanded form of the LHS. Further, it can appear only once since it only makes one pair of columns dependent. Similar reasoning for $b$ and $c$ tells you that $b$ and $c$ must also be factors of the RHS.

Also note that if $a+b+c = 0$, the determinant becomes:

$$\begin{vmatrix} a^2 & a^2 & a^2\\ b^2&b^2&b^2\\c^2&c^2&b^2 \end{vmatrix} = 0$$

(by expressing $a+b$ as $-c$ and so on.) Now in this matrix three pairs of columns are dependent (pair-wise, of course), and so the factor $(a+b+c)$ must appear thrice on the RHS.

Now by mentally expanding the determinant with your focus on the exponent of the terms and not the coefficients, etc., you will see that all the exponents match, i.e. you are not missing any factors. So you know the RHS is a constant multiple of $abc(a+b+c)^3$. Setting $a=-1, b=c=1$ and expanding you will get that the constant coefficient is $2$ by comparison.

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\begin{vmatrix}(b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2\end{vmatrix}

Employing $C_1'=C_1-C_3,C_2'=C_2-C_3$

=\begin{vmatrix}(b+c)^2-a^2 & a^2-a^2 & a^2 \\ b^2-b^2 & (c+a)^2-b^2 & b^2 \\ c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2\end{vmatrix}

=\begin{vmatrix}(b+c+a)(b+c-a) & 0 & a^2 \\ 0 & (c+a+b)(c+a-b) & b^2 \\ (c-a-b)(c+a+b) & (c+a+b)(c-a-b) & (a+b)^2\end{vmatrix}

$=(a+b+c)^2\cdot$\begin{vmatrix}b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ c-a-b & c-a-b & (a+b)^2\end{vmatrix}

$=\frac12\cdot(a+b+c)^2\cdot$\begin{vmatrix}b+c-a & 0 & 2a^2 \\ 0 & c+a-b & 2b^2 \\ c-a-b & c-a-b & 2(a+b)^2\end{vmatrix}

Employing $C_3'=aC_1+bC_2+C_3$

$=\frac12\cdot(a+b+c)^2\cdot$\begin{vmatrix}b+c-a & 0 & a(a+b+c) \\ 0 & c+a-b & b(a+b+c) \\ c-a-b & c-a-b & 2(a+b)(a+b+c)\end{vmatrix}

$=\frac12\cdot(a+b+c)^3\cdot$\begin{vmatrix}b+c-a & 0 & a \\ 0 & c+a-b & b \\ c-a-b & c-a-b & 2(a+b)\end{vmatrix}

Can you take it home from here?

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