Question

2x divide by 7 minus 3 Y / 8 whole cube

Answer 1

Apply the formula
(a-b)^3

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$( \frac{2x}{7} - \frac{3y}{8} ) {}^{3} = ( \frac{2x}{7}) {}^{3} - ( \frac{3y}{8} ) {}^{3} - 3 \times \frac{2x}{7} \times \frac{3y}{8}( \frac{2x}{7} - \frac{3y}{8}) \\ \\ = \frac{8 {x}^{3} }{343} - \frac{27 {y}^{3} }{512} - \frac{18xy}{56} ( \frac{16x - 21y}{56}) \\ \\ = \frac{8 {x}^{3} }{343} - \frac{27 {y}^{3} }{512} - \frac{ 9xy}{28} ( \frac{16x - 21y}{56} ) \: \:ans$