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Algebraic Identities

Simplify: a...

Question

Simplify: ${(\frac{a}{2} - \frac{2}{3} b)}^{3} - {(\frac{a}{2} + \frac{2}{3} b)}^{3}$

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Solution

${(\frac{a}{2} - \frac{2 b}{3})}^{3} - {(\frac{a}{2} + \frac{2 b}{3})}^{3} = [(\frac{a}{2} - \frac{2 b}{3}) - (\frac{a}{2} + \frac{2 b}{3})] [{(\frac{a}{2} - \frac{2 b}{3})}^{2} + {(\frac{a}{2} + \frac{2 b}{3})}^{2} + (\frac{a}{2} - \frac{2 b}{3}) (\frac{a}{2} + \frac{2 b}{3})] = [(\frac{a}{2} - \frac{2 b}{3} - \frac{a}{2} - \frac{2 b}{3})] [2 {{(\frac{a}{2})}^{2} + {(\frac{2 b}{3})}^{2}} + (\frac{a^{2}}{4} - \frac{4 b^{2}}{9})] = [\frac{- 4 b}{3}] [2 (\frac{a^{2}}{4} + \frac{4 b^{2}}{9}) + (\frac{a^{2}}{4} - \frac{4 b^{2}}{9})] = [\frac{- 4 b}{3}] [\frac{2 a^{2}}{4} + \frac{8 b^{2}}{9} + \frac{a^{2}}{4} - \frac{4 b^{2}}{9}] = [\frac{- 4 b}{3}] [\frac{3 a^{2}}{4} + \frac{4 b^{2}}{9}] = \frac{- 12 a^{2} b}{12} - \frac{16 b^{3}}{27} = - b (a^{2} + \frac{16 b^{2}}{27})$

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