Question 93260
2as= 2uv-2u^2+v^2-2vu+u^2

apparantly by collecting terms this is equal to 2as=v^2-u^2, could someone please explain this last step for me
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2uv-2u^2+v^2-2vu+u^2

=2u(v-u) + [v^2-2uv+u^2]

= 2u(v-u) + [v-u]^2

=(v-u)[2u + v-u]

= (v-u)(v+u)

= v^2-u^2

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Cheers,
Stan H.