Answer:
[tex]\large\boxed{(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz}[/tex]
Step-by-step explanation:
Why?
Let y + z = t. Then
[tex]x+y+z=x+t\to(x+y+z)^2=(x+t)^2[/tex]
Use
[tex](a+b)^2=a^2+2ab+b^2\qquad(*)[/tex]
[tex](x+t)^2=x^2+2xt+t^2[/tex]
Come back to substitution:
[tex]x^2+2x(y+z)+(y+z)^2[/tex]
Use (*) and the distributive property: a(b + c) = ab + ac
[tex]x^2+2xy+2xz+y^2+2yz+z^2\\\\x^2+y^2+z^2+2xy+2xz+2yz[/tex]
