Question 1037399
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If x-y=4 and xy=21 find x^3-y^3
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<pre>
{{{x^3-y^3}}} = {{{(x-y)*(x^2 + xy + y^2)}}}   (which is classic identity)

= {{{(x-y)*((x-y)^2 + 3xy)}}}   (now plug-in the given data)

= 4*(4^2 +3*21) = 4*(16+63) = 4*79 = 316.
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