hello, plugging $c=\pm\frac{1}{2}\sqrt{2(a^2+b^2)}$ in the given term and simplifying we get $3$ in both cases. Sonnhard.

$2a^2 - b^2 - c^2 = 2a^2 - (2c^2-a^2) - c^2 = 3(a^2-c^2) = 3(a-c)(a+c)$. $(a-c)(a+b+2c) = a^2 - c^2 + (a-c)(b+c) =$ $ a^2 - 2c^2 -bc + a(b+c) =$ $ -b^2 -bc + a(b+c) = -b(b+c) + a(b+c) =$ $ (a-b)(b+c)$. So $(a+b+2c)(2a^2 - b^2 - c^2) = 3(a-b)(a+c)(b+c)$.

Good but pretty straightforward nonetheless. Note that $a^2+b^2=2c^2 \implies a^2-c^2=c^2-b^2=(a-c)(a+c)=(b+c)(c-b)$ So $\frac{(a+b+2c)(2a^2-b^2-c^2)}{(a-b)(a+c)(b+c)}=\frac{3(a+b+2c)(a-c)(a+c)}{(a-b)(a+c)(b+c)}=\frac{3(a+b+2c)(a-c)}{(a-b)(b+c)}=\frac{3(a+b+2c)}{a-b} \cdot \frac{c-b}{a+c}=3\frac{(a+c)+(b+c)}{a+c}\cdot \frac{c-b}{a-b}=3(1+\frac{b+c}{a+c}) \cdot \frac{c-b}{a-b}=3(1+\frac{a-c}{c-b}) \cdot \frac{c-b}{a-b}=3 \cdot \frac{a-b}{c-b} \cdot \frac{c-b}{a-b}=3$ which is an integer of course