How do you expand ${(x - y)}^{3}$ $(x-y)^3$ ?

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Answer 1 · 2018-04-10T16:36:39

= $x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3}$ $x^3-3x^2y+3xy^2-y^3$

$(x - y) (x - y) . = x^{2} - x y - x y + y^{2}$ $(x-y)(x-y). = x^2-xy-xy+y^2$

= $x^{2} - 2 x y + y^{2}$ $x^2-2xy+y^2$

= $(x^{2} - 2 x y + y^{2}) (x - y)$ $(x^2-2xy+y^2)(x-y)$

= $x^{3} - x^{2} y - 2 x^{2} y + 2 x y^{2} + x y^{2} - y^{3}$ $x^3-x^2y-2x^2y+2xy^2+xy^2-y^3$

= $x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3}$ $x^3-3x^2y+3xy^2-y^3$

Answer 2 · 2018-05-08T15:47:29

$x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}$ $x^3-y^3-3x^2y+3xy^2$

${(x - y)}^{3} = (x - y) (x - y) (x - y)$ $(x-y)^3=(x-y)(x-y)(x-y)$

Expand the first two brackets:

$(x - y) (x - y) = x^{2} - x y - x y + y^{2}$ $(x-y)(x-y)=x^2-xy-xy+y^2$

$\Rightarrow x^{2} + y^{2} - 2 x y$ $rArr x^2+y^2-2xy$

Multiply the result by the last two brackets:

$(x^{2} + y^{2} - 2 x y) (x - y) = x^{3} - x^{2} y + x y^{2} - y^{3} - 2 x^{2} y + 2 x y^{2}$ $(x^2+y^2-2xy)(x-y)=x^3-x^2y+xy^2-y^3-2x^2y+2xy^2$

$\Rightarrow x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}$ $rArr x^3-y^3-3x^2y+3xy^2$

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.

How do you expand (x−y)3(x-y)^3?