An algebra problem by Rohith M.Athreya

$(x+y+z)^{2} +xy+(y+z)^{2} +z^{2}+2xz+zy$

consider the $xy$ and one $xz$ term and group them

$(x+y+z)^{2} +x(y+z)+(y+z)^{2} +z^{2}+xz+zy$

group second and third term

$(x+y+z)^{2} +(y+z)(x+y+z) +z^{2}+xz+zy$

$(x+y+z)^{2} +(y+z)(x+y+z) +z(x+y+z)$

factorise again to get

$(x+y+z)(x+2y+3z)$

this is simply zero!! $\int log rx \, dx$

Given that $x+2y+3z = 0$ $\implies \color{#3D99F6}{x = -2y - 3z}$ , also that $\color{#D61F06}{xy=4}$ . Now we have:

$\begin{aligned} X & = (\color{#3D99F6}{x}+y+z)^2 +\color{#D61F06}{xy}+(y+z)^2 + z^2 +2z\color{#3D99F6}{x} + yz \\ & = (\color{#3D99F6}{-2y-3z}+y+z)^2 +\color{#D61F06}{4}+(y+z)^2 + z^2 +2z(\color{#3D99F6}{-2y-3z}) + yz \\ & = (-y-2z)^2 +4+y^2+2yz+z^2 + z^2 -4yz-6z^2 + yz \\ & = y^2+4yz + 4z^2 +4+y^2+2yz+z^2 + z^2 -4yz-6z^2 + yz \\ & = 2y^2 + 3yz + 4 \\ & = y(\color{#3D99F6}{2y+3z}) + 4 \\ & = y(\color{#3D99F6}{-x}) + 4 \\ & = -\color{#D61F06}{xy} + 4 \\ & = -\color{#D61F06}{4} + 4 \\ & = \boxed{0} \end{aligned}$