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Question Statment:- Show that \begin{align*} \begin{vmatrix} (a+b)^2 & ca & bc \\ ca & (b+c)^2 & ab \\ bc & ab & (c+a)^2 \\ \end{vmatrix} =2abc(a+b+c)^3 \end{align*}

My Attempt:-

$$\begin{aligned} &\begin{vmatrix} \\(a+b)^2 & ca & bc \\ \\ca & (b+c)^2 & ab \\ \\bc & ab & (c+a)^2 \\\ \end{vmatrix}\\ =&\begin{vmatrix} \\a^2+b^2+2ab & ca & bc \\ \\ca & b^2+c^2+2bc & ab \\ \\bc & ab & c^2+a^2+2ac \\\ \end{vmatrix}\\ =&\dfrac{1}{abc}\begin{vmatrix} \\ca^2+cb^2+2abc & ca^2 & b^2c \\ \\ac^2 & ab^2+ac^2+2abc & ab^2 \\ \\bc^2 & a^2b & bc^2+a^2b+2abc \\\ \end{vmatrix}\\\\ &\qquad (C_1\rightarrow cC_1, C_2\rightarrow aC_2, C_3\rightarrow bC_3)\\\\ =&\dfrac{2}{abc}\times\begin{vmatrix} \\ca^2+cb^2+abc & ca^2 & b^2c \\ \\ab^2+ac^2+abc & ab^2+ac^2+2abc & ab^2 \\ \\bc^2+a^2b+abc & a^2b & bc^2+a^2b+2abc \\\ \end{vmatrix}\\\\ &\qquad (C_1\rightarrow C_1+C_2+C_3)\\\\ =&\dfrac{2abc}{abc}\left(\begin{vmatrix} \\a^2+b^2 & a^2 & b^2 \\ \\b^2+c^2 & b^2+c^2+2bc & b^2 \\ \\c^2+a^2 & a^2 & c^2+a^2+2ac \\\ \end{vmatrix}+ \begin{vmatrix} \\1 & ca^2 & b^2c \\ \\1 & ab^2+ac^2+2abc & ab^2 \\ \\1 & a^2b & bc^2+a^2b+2abc \\\ \end{vmatrix}\right) \end{aligned}$$

The second determinant in the last step can be simplified to \begin{vmatrix} \\1 & ca^2 & b^2c \\ \\0 & ab^2+ac^2+2abc-ca^2 & ab^2-b^2c \\ \\0 & a^2b-ca^2 & bc^2+a^2b+2abc-b^2c \\\ \end{vmatrix}

I couldn't proceed further with this, so your help will be appreciated and if any other simpler way is possible please do post it too.

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One can use factor theorem to get a simpler solution. If we put $a=0$, we get \begin{align*} \begin{vmatrix} (a+b)^2 & ca & bc \\ ca & (b+c)^2 & ab \\ bc & ab & (c+a)^2 \\ \end{vmatrix} =\begin{vmatrix} b^2 & 0 & bc \\ 0 & (b+c)^2 & 0 \\ bc & 0 & c^2 \\ \end{vmatrix} = 0 \end{align*} Hence $a$ is a factor. Similarly $b, c$ are factors. Again, put $a+b+c=0$, we get \begin{align*} \begin{vmatrix} c^2 & ca & bc \\ ca & a^2 & ab \\ bc & ab & b^2 \\ \end{vmatrix} =abc\begin{vmatrix} c & a & b \\ c & a & b \\ c& a & b \\ \end{vmatrix} \end{align*} Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$. Putting $a=b=c=1$, we obtain \begin{align*} 27k = \begin{vmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \\ \end{vmatrix} =54 \end{align*} and $k=2$. Thus the given determinant equals $2abc(a+b+c)^3$

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  • $\begingroup$ Why is it that when you put $a+b+c=0$ and as all rows are identical then you concluded that $(a+b+c)^2$ is also a factor. $\endgroup$ – user350331 Aug 12 '16 at 6:39
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    $\begingroup$ Suppose that in a determinant the entries are polynomials in $x$. Suppose also that when we put $x=a$, two rows, say $R_1, R_2$ become identical. This means that when we apply $R_1 \rightarrow R_1 - R_2$ and then put $x=a$, $R_1 = 0$. That is, $x-a$ is a factor of $R_1-R_2$ and can be taken out. When $R_1 = R_2=R_3$, the same argument shows that $x-a$ divides both $R_1 - R_2$ and $R_2-R_3$ and hence $(x-a)^2$ can be taken out. $\endgroup$ – user348749 Aug 12 '16 at 6:44
  • $\begingroup$ That was a wonderful explanation, my other question is how that you conclude that the last remaining linear factor would be $k(a+b+c)$, and also what do you refer as symmetric in your solution. $\endgroup$ – user350331 Aug 12 '16 at 6:59
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    $\begingroup$ Note that the determinant does not change value when you make the following changes: $a \leftrightarrow b$ or $b \leftrightarrow c$ or $c \leftrightarrow a$. Hence the determinant is invariant under $S_3$. That means that all its factors are also symmetric polynomials. A symmetric polynomial of degree 1 in $a,b,c$ must remain invariant under the above transformations and hence the coefficients of $a,b,c$ must be the same. Thus it must be of the form $k(a+b+c)$ $\endgroup$ – user348749 Aug 12 '16 at 7:07
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    $\begingroup$ $S_3$ is the symmetric group on three symbols. $\endgroup$ – user348749 Aug 12 '16 at 7:13
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Let me try. You have $$LHS = [(a+b)(b+c)(c+a)]^2 + 2(abc)^2 - \sum b^2c^2(b+c)^2.$$

Note that $(a+b)(b+c)(c+a) = \sum bc(b+c) + 2abc$. Then, $$LHS = \left(\sum bc(b+c)\right)^2 + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right) - \sum b^2c^2(b+c)^2$$

$$=2\sum a^2bc(a+b)(a+c) + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right)$$

$$ =2abc\left(\sum a(a+b)(a+c) + 3abc + 2\sum bc(b+c)\right) $$

$$ = 2abc \left(\sum a^3 + 6abc + 3\sum bc(b+c)\right) $$

$$ = 2abc \left(a+b+c\right)^3 $$

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  • $\begingroup$ How did you arrive at that LHS in the beggining $\endgroup$ – user350331 Aug 12 '16 at 5:09

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