<s>[INST] simple method for expanding binomial with 3 or more terms? I've seen
Binomial Theorem Question (Expansion of Three Terms)
Binomial Theorem with Three Terms
Expanding Equation with Binomial Theorem
but I'm not such a math expert, I need things explained in simple terms.
Basically I've heard that the solve
(x + y)^

it's essentially
(x + y)(x + y)

and then multiply the first terms, (leftmost), then first term of one with last term of the other (outer) then second term of one with first term of other (inner) then the two last terms of each (right most), and add them up, so
x * x + x * y + y * x + y * y =

x^2 + 2xy + y^2

that is pretty much all I know, now if I want to solve a more complicated binomial, with three or more terms, for example
(x + y + z)(x + y + z)

would I use a similar method? Meaning do I start with the left most terms, then x * y, then, what? then do I do x * z and then move on to the next right term, y, and do y * x + y * y + y * z, and then do the same for z, meaning z * x + z * y + z * z ? Am I missing something here, or is that it?
 [/INST] You do not need any theorems or results other than the distributive property: $a(b+c)=ab+ac$. Applying this once, we have $(x+y)(x+y)=x(x+y)+y(x+y)$. Here, the left $x+y$ is playing the role of $a$, and the right $x$ and $y$ are playing the roles of $b$ and $c$ respectively. Then we can use the same property again to see $x(x+y)=x^2+xy$ and $y(x+y)=yx+y^2$. Since $xy=yx$, we have
\begin{align}
(x+y)(x+y)&=x(x+y)+y(x+y)\\
&=x^2+xy+yx+y^2\\
&=x^2+xy+xy+y^2\\
&=x^2+2xy+y^2.
\end{align}
For three terms we can do the same thing:
\begin{align}
(x+y+z)(x+y+z)&=x(x+y+z)+y(x+y+z)+z(x+y+z)\\
&=x^2+xy+xz+xy+y^2+yz+xz+yz+z^2\\
&=x^2+y^2+z^2+2xy+2xz+2yz.
\end{align}
</s>