show that determinant
$$\left|\matrix{
x^2+L & xy & xz \\
xy & y^2+L & yz \\
xz & yz & z^2+L \\
}\right|
= L^2(x^2+y^2+z^2+L)$$
without expanding by using the appropriate properties of determinant.
All i can do is LHS
$$x^2y^2z^2\left|\matrix{
1+L/x^2 & 1 & 1 \\
1 & 1+L/y^2 & 1 \\
1 & 1 & 1+L/z^2 \\
}\right|$$
Best Answer
Look at this form of your matrix $$\pmatrix{L & 0 & 0 \\ 0 & L & 0 \\ 0 & 0 & L\\} + \pmatrix{x & 0 & 0 \\ 0 & y & 0\\ 0 & 0 & z}\pmatrix{ x & y & z \\ x & y & z \\ x & y & z \\}$$
and consider your right eigenvector $\pmatrix{x \\ y \\ z}$ with eignevalue $ x^2 + y^2 + z^2 + L$
I suspect you may not yet be familiar with eigenvalues, so if that is true, maybe this is some good motivation to learn.
Eigenvector review:
A right eigenvector $\mathbf{v}$ with eigenvalue $\lambda$ is defined for a matrix $\mathbf{A}$ as $$\mathbf{A}\mathbf{v} = \mathbf{v}\lambda$$ Here we see that $$\pmatrix{x^2 + L & xy & xz \\ xy & y^2 + L & yz \\ xz & yz & z^2 + L}\pmatrix{x \\ y \\ z} = \pmatrix{x \\ y \\ z}(x^2 + y^2 + z^2 + L)$$ In this case all other eigenvectors come from the two dimensional subspace defined as the space of vectors orthogonal to the first eigenvector, and they have eignevalue $L$ (this can be seen more easily from the first form I stated for you matrix).
The unit hypercube of hypervolume one with edges defined by those eigenvectors then is transformed by your matrix to have volume $(x^2 + y^2 + z^2 + L)L^2$ and that is the determinant for your matrix.