Without removing brackets at any stage factorize $$2y(y+z) - (x+y)(x+z)$$

Regard the expression as a quadratic in $$x$$. Putting $$x = y$$ gives $$2y(y+z) - 2y(y+z) = 0$$, so by the factor theorem $$(y - x)$$ is a factor, and since the coefficient of $$x^2$$ is $$-1$$ the expression equals $$(y - x)(x + L)$$ for some $$L$$ not involving $$x$$. Putting $$x = 0$$ gives $$2y(y+z) - y \cdot z = y(2y + z)$$ on one side and $$y \cdot L$$ on the other, so $$L = 2y + z$$. Hence $$2y(y+z) - (x+y)(x+z) = (y - x)(x + 2y + z)$$.