Find the constant $k$ from the determinant

Question

Given: $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.

If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.

Answer 1

Let $$a=b=c=1$$ and you get the matrix

$$\begin{vmatrix}4&1&1\\1 &4&1 \\1&1& 4\end{vmatrix}=27k$$

The determinant is easily evaluated to be $54$ so $$27k=54$$.

Thus $$k=2$$

Answer 2

It is also simple to compute the posted determinant, $D$ for short, thus ignoring the information on the special form of the result. We are subtracting at the first step the first column from the other two. $$ \begin{aligned} D &= \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \\[2mm] &= \begin{vmatrix} (b+c)^2 & a^2- (b+c)^2 & a^2 - (b+c)^2 \\ b^2 & (c+a)^2-b^2 & 0 \\ c^2 & 0 & (a+b)^2-c^2 \end{vmatrix} \\[2mm] &= \begin{vmatrix} (b+c)^2 & a- b-c & a- b-c \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \cdot(a+b+c)^2 \\[2mm] &= \begin{vmatrix} 2bc & -2c & - 2b \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \cdot(a+b+c)^2 \\[2mm] &= \begin{vmatrix} 2bc-bc-bc & -2c & - 2b \\ b^2+\frac b2(c+a-b)+\frac c2\cdot 0 & c+a-b & 0 \\ c^2+\frac c2\cdot 0+\frac c2(a+b-c) & 0 & a+b-c \end{vmatrix} \cdot(a+b+c)^2 \\[2mm] &= \begin{vmatrix} 0 & -2c & - 2b \\ \frac b2 & c+a-b & 0 \\ \frac c2 & 0 & a+b-c \end{vmatrix} \cdot(a+b+c)^3 \\[2mm] &= -\begin{vmatrix} 0 & c & b \\ b & c+a-b & 0 \\ c & 0 & a+b-c \end{vmatrix} \cdot(a+b+c)^3 \\[2mm] &=2abc(a+b+c)^3\ . \end{aligned} $$ (At the last step, using Sarrus, only two out of six terms survive, we can factor $bc$, and $-(a+b-c)-(c+a-b)=-2a$ leads immediately to the result. No step was omitted, symmetry was used when possible.)

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Computer check:

sage: var('a,b,c');
sage: A = matrix(3,3, [(b+c)^2, a^2, a^2, b^2, (a+c)^2, b^2, c^2, c^2, (a+b)^2] )
sage: A.det().factor()
2*(a + b + c)^3*a*b*c

Answer 3

It's not long: $$\prod_{cyc}(a+b)^2+2a^2b^2c^2-\sum_{cyc}a^2b^2(a+b)^2=$$ $$\sum_{cyc}\left(a^4b^2+a^4c^2+2a^3b^3+2a^4bc+2a^3b^2c+2a^3c^2b+2a^2b^2c^2+4a^3b^2c+4a^3c^2b+\tfrac{4}{3}a^2b^2c^2\right)+$$ $$+\sum_{cyc}\left(\frac{2}{3}a^2b^2c^2-a^4b^2-a^4c^2-2a^3b^3\right)=$$ $$=\sum_{cyc}(2a^4bc+6a^3b^2c+6a^3c^2b+4a^2b^2c^2)=2abc(a+b+c)^3.$$ Two minutes of computations! By hand, of course.