Let $B$ be a $3\times3$ matrix, and let $f(x,y)= \det\left(xB+yB^T\right)$. Show that $\det\left(xB+yB^T\right)$ is a multiple of $x+y$, where $x,y$ are any real numbers and $B$ is any $3 \times 3$ matrix with real entries.
I can only think a tedious method to show this question by assuming a determinant to do it. Can anyone give me a idea or method to finish it easily?
Let $f(x,y) = \det(xB + yB^T)$. Then $f(x, -x) = \det(xB - xB^T) = x^3 \det(B- B^T)$
The matrix $A= B-B^T$ is an odd-size skew-symmetric $(A^T = -A)$ matrix. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^3 \det(A) = -\det(A)$, it follows that $\det(A) = 0$.
Added: $f(x,y) \in \mathbb{R}[x,y] = (\mathbb{R}[x])[y]$ is a polynomial, hence $f(x,y) = (x+y)g(x,y) + h(x)$. Therefore $x+y$ divides $f(x,y)$ in $\mathbb{R}[x,y]$ if and only if $h(x) = 0$, and the latter is equivalent to $f(x, -x) = 0$.