How do you expand (x−y)3?

Explanation:
${\displaystyle (x-y)(x-y)=x^2-xy-xy+y^2}$
${\displaystyle =x^2-2xy+y^2}$
${\displaystyle =(x^2-2xy+y^2)(x-y)}$
${\displaystyle x^3-x^{2}y-2x^{2}y+2x{y}^2+x{y}^2-y^3}$
${\displaystyle =x^3-3x^{2}y+3x{y}^2-y^3}$

Explanation:
${\displaystyle {(x-y)}^3=(x-y)(x-y)(x-y)}$
Expand the first two brackets:
${\displaystyle (x-y)(x-y)=x^2-xy-xy+y^2}$
${\displaystyle \Rightarrow x^2+y^2-2xy}$
Multiply the result by the last two brackets:
${\displaystyle (x^2+y^2-2xy)(x-y)=x^3-x^{2}y+x{y}^2-y^3-2x^{2}y+2x{y}^2}$
${\displaystyle \Rightarrow x^3-y^3-3x^{2}y+3x{y}^2}$
Always expand each term in the bracket by all the other terms in the other brackets, but never multiply two or more terms in the same bracket.

Explanation

The expression $(x-y{)}^3$ can be written as, $(x-y)(x-y)(x-y)$
First simplify (x-y)(x-y)by binomial multiplication.
$(x-y)(x-y)=x^2-2xy+y^2$
Now multiply (x-y) with $x^2-2xy+y^2$
$(x-y)(x-y)(x-y)=(x-y)(x^2-2xy+y^2)$
$=x^3-2x^{2}y+x{y}^2-y{x}^2+2x{y}^2-y^3$
$=x^3-3x^{2}y+3x{y}^2-y^3$
Thus, the expansion of $(x-y{)}^{3}is{x}^3-y^3-3x^{2}y+3x{y}^2$