Prove $\det(P+Q+R)=\det(P+Q)+\det(Q+R)+\det(R+P)-\det(P)-\det(Q)-\det(R)$

Notice for any $A \in M_2(\mathbb{C})$, the characteristic polynomial has the form

$$\chi_A(\lambda) = \det(\lambda I_2 - A) = \lambda^2 - {\rm tr}(A)\lambda + \det(A)$$

Using Cayley Hamilton and taking trace again, we can express the determinant in terms of the trace:

$$A^2 - {\rm tr}(A) A + \det(A) I_2 = 0 \quad\implies\quad \det(A) = \frac12\left({\rm tr}(A)^2 - {\rm tr}(A^2)\right) $$ For any $A, B \in M_2(\mathbb{C})$, define a bracket between them by $$\langle A, B \rangle \stackrel{def}{=} \det(A+B) - \det(A) - \det(B)$$ Using above expression of determinant, it is easy to verify:

$$\langle A, B \rangle = {\rm tr}(A){\rm tr}(B) - {\rm tr}(AB)$$

This means this sort of bracket is bilinear in terms of its arguments. As a result, we obtain

$$\begin{align}\det(P+Q+R) = & \det(P+Q) + \det(R) + \langle P+Q, R \rangle\\ = & \det(P+Q) + \det(R) + \color{red}{\langle P, R \rangle} + \color{blue}{\langle Q, R \rangle}\\ = & \det(P+Q) + \det(R) +\left(\color{red}{\det(P+R) - \det(P) - \det(R)}\right)\\ & \quad + \left(\color{blue}{\det(Q+R) - \det(Q) - \det(R)}\right)\\ = & \det(P+Q) + \det(P+R) + \det(Q+R) - \det(P) - \det(Q) - \det(R) \end{align} $$

HINT:

$\det S$ is a homogenous polynomial of degree $2$ in the entries of $S$, that is, a quadratic form. It is easier to prove the general fact: If $q\colon V \to W$ is a quadratic form then $$q(P+Q+R) - q(P+Q)- q(P+R) -q(Q+R) + q(P) + q(Q) + q(R)=0$$

The thing about quadratic forms, they come from bilinear forms ( symmetric too, if $char \ne 2$, but that is not important for this). So $$q(S) = b(S,S)$$ Substitute in the above, use bilinearity, and observe that all of the terms cancel out.

$\bf{Added:}$ One should try to guess what is the equality that holds for cubic forms, or, more generally, for forms of degree $n$. It is also related to "polarization".

If $\det(P) = p_{11}p_{22}-p_{21}p_{12}$, thrn

$$\det(P+Q+R) = (p_{11}+q_{11}+r_{11})(p_{22}+q_{22}+r_{22})-(p_{21}+q_{21}+r_{21})(p_{12}+q_{12}+r_{12})$$

You can start in that way.