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Using the property of determinant, verify that: $$\left| \begin{matrix} (b+c)^2&a^2&a^2 \\ b^2&(c+a)^2&b^2 \\ c^2&c^2&(a+b)^2 \\ \end{matrix}\right|=2abc(a+b+c)^3$$

My Attempt: $$L.H.S= \left| \begin{matrix} (b+c)^2&a^2&a^2 \\ b^2&(c+a)^2&b^2 \\ c^2&c^2&(a+b)^2 \\ \end{matrix}\right|$$ $$C_1\to C_1-C_2 \textrm{and} C_2\to C_2-C_3$$ $$= \left| \begin{matrix} (b+c)^2-a^2&a^2-a^2&a^2 \\ b^2-(c+a)^2&(c+a)^2-b^2&b^2 \\ c^2-c^2&c^2-(a+b)^2&(a+b)^2 \\ \end{matrix}\right|$$ $$= \left| \begin{matrix} (a+b+c)(b+c-a)&0&a^2 \\ (a+b+c)(b-c-a)&(a+b+c)(c+a-b)&b^2 \\ 0&(a+b+c)(c-a-b)&(a+b)^2 \\ \end{matrix}\right|$$ Taking common $(a+b+c)$ from $C_1$ and $C_2$, $$=(a+b+c)^2 \left| \begin{matrix} b+c-a&0&a^2 \\ b-c-a&c+a-b&b^2 \\ 0&c-a-b&(a+b)^2 \\ \end{matrix}\right|$$

How do I get the proof?

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  • $\begingroup$ Keep going, expand remaining determinant. $\endgroup$
    – coffeemath
    Apr 29, 2018 at 6:06
  • $\begingroup$ @coffeemath, can't further simplification be done before expanding? $\endgroup$
    – pi-π
    Apr 29, 2018 at 6:08

1 Answer 1

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If $a=0$ or $b=0$, or $c=0$, the result clearly holds. I will assume $a,b,c$ are all non-zero.

I will pick up from where you left off.

\begin{align} &\begin{vmatrix} b+c-a & 0 & a^2 \\ b-c-a & c+a-b & b^2 \\ 0 & c-a-b & (a+b)^2 \end{vmatrix}, & R_3 \to R_3-(R_1+R_2)\\ &=\begin{vmatrix} b+c-a & 0 & a^2 \\ b-c-a & c+a-b & b^2 \\ -2(b-a) & -2a & 2ab \end{vmatrix}, & C_1 \to C_1+C_2\\ &=\begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ -2b & -2a & 2ab \end{vmatrix} \\ &=\frac1{ab}\begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ 0 & bc+ab-b^2 & b^2 \\ -2ab & -2ab & 2ab \end{vmatrix} \\ &=2\begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ 0 & bc+ab-b^2 & b^2 \\ -1 & -1 & 1 \end{vmatrix}, & C_1 \to C_1+C_3, C_2 \to C_2 + C_3\\ &=2\begin{vmatrix} ab+ac & a^2 & a^2 \\ b^2 & bc+ab & b^2 \\ 0 & 0 & 1 \end{vmatrix} \\ &=2\begin{vmatrix} ab+ac & a^2 \\ b^2 & bc+ab \end{vmatrix} \\ &= 2ab\begin{vmatrix} b+c & a \\ b & a+c \end{vmatrix} \\ &=2ab((b+c)(a+c)-ab)\\ &= 2ab(c^2+(a+b)c)\\ &=2abc(a+b+c) \end{align}

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