5
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $2abc(a+b+c)^3$

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $S_3$ is the symmetric group on three symbols. $\endgroup$ – user348749 Aug 12 '16 at 7:13
0
$\begingroup$

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

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

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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.