<s>[INST] Proving $(x^2 - y^2) = (x + y)(x-y)$ I am trying to prove from the field axioms that $(x^2 - y^2) = (x+y)(x-y)$. I am going to take for granted the definition of subtraction as the addition of a negation and that $x(-y) = (-x)y = -(xy)$. Here is my attempt.
We have:
\begin{align*}
(x+y)(x-y) & = x(x-y) + y(x-y) & & \text{distributive law} \\
& = x(x + (-y)) + y(x + (-y)) & & \text{definition of subtraction} \\
& = (xx + x(-y)) + (yx + y(-y)) & & \text{distributive law} \\
& = (x^2 + (-(xy))) + (yx + (-y^2)) & & \text{uniqueness of additive inverse; definition of square} \\
& = (x^2 + (-(yx))) + (yx + (-y^2)) & & \text{commutativity of multiplication} \\
& = (x^2 + \left(-(yx) + yx \right) + (-y^2) & & \text{associativity of addition} \\
& = (x^2 + 0) + (-y^2) & & \text{additive inverse axiom} \\
& = x^2 + (-y^2) & & \text{additive identity axiom} \\
& = x^2 - y^2 & & \text{definition of subtraction} 
\end{align*}
How does this look?
 [/INST] Another way to approach it for the sake of curiosity:
\begin{align*}
x^{2} - y^{2} & = (x^{2} - xy) + (xy - y^{2})\\\\
& = x(x-y) + y(x - y)\\\\
& = (x+y)(x-y)
\end{align*}
Hopefully this helps!
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