<s>[INST] Computing determinant of a specific matrix: This question might seem very easy to some. But I am having a very tough time solving it.
$$
A = \begin{pmatrix}
1 + x^2 - y^2 - z^2& 2(xy + z) & 2(zx-y)  \\ 
2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x)  \\
2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2 
\end{pmatrix}
$$
Then $ \det A$ is: 
(a) $(1 + xy + yz + zx)^3$
(b) $(xy + yz + zx)^3$
(c) $(1 + x^2 + y^2 + z^2)^3$
(d) $(1 + x^3 + y^3 + z^3)^2$
I am interested in an approach other than the brute-force approach of expanding the  $3 \times 3$ determinant. I tried row and column operations, but I wasn't able to see a simplification. I couldn't write A as product of 2 matrices either, so I am clueless. (Also I want a rigorous proof, not methods like put $(x,y,z) = (0,1,2) $ etc. and eliminate 3 options to get the right answer.)
 [/INST] Let $B = \begin{bmatrix}0 & z & -y\\ -z & 0 & x \\ y & -x & 0\end{bmatrix}$ be half of the anti-symmetric part of $A$. Since its square
$$B^2 = \begin{bmatrix}-y^2-z^2 & xy & xz\\ xy & -x^2-z^2 & yz\\xz & yz & - x^2-y^2\end{bmatrix}$$
has same off-diagonal elements as half of the symmetry part of $A$, we obtain following decomposition of $A$.
$$A = (1+r^2) I_3 + 2B(I+B)$$
It is sort of well known $B$ has eigenvalues $0, \pm ir$ where $r = \sqrt{x^2+y^2+z^2}$.
This means $A$ has eigenvalues $$1+r^2 \quad\text{and}\quad 1+r^2 \pm 2ir(1 \pm ir) = 1-r^2 \pm 2ir = (1 \pm ir)^2$$
and hence
$$\det(A) = (1+r^2)(1+ir)^2(1-ir)^2 = (1+r^2)^3 = (1+x^2+y^2+z^2)^3$$
</s>