Let A be an 3 x 3 matrix that contains complex numbers, and $A^t$ its transpose. Considering that
$\det(A+A^t)=8 $ $ $ and $ $ $\det(A+2A^t)=27$ $ $ prove that $ $ $ \det(A)=1 $
and that $ $ $ \det( xA+yA^t) = (x+y)^3 $ $ $ for all $ $ $x,y\in \mathbb{C}$.
I need a hint to solve this problem. Thank you in advance!
Consider the function $g(x)=\det((\frac 12 - x)A+(\frac 12 +x) A^t)$. This is a polynomial of variable $x$ and the degree of $g$ is at most three. The given conditions rewrite as $g(0)=1$ and $g(\frac 16)=1$. Also note that $$\left((\frac 12 - x)A+(\frac 12 +x) A^t\right)^t = (\frac 12 + x)A+(\frac 12 - x) A^t.$$ Applying $\det$ and using $\det X = \det X^t$ we get that $g(x)=g(-x)$ for all $x$.
So, $g$ is an even polynomial of degree at most three satisfying $g(0)=g(\frac 16)=1$. These conditions already imply that $g(x)=1$ for all $x$. I'll leave it to you as an exercise. Then it follows immediately that $\det(A)=g(-\frac 12)=1$. It is also quite easy to prove that $\det(xA+yA^t)=(x+y)^3$, just consider $g(\frac{y-x}{2(x+y)})$.