Question

(3y-1/3x)³=





допоможіть будь-ласка​

Answer 1

Ответ:

${(3y - \frac{1}{3x} )}^{3} = \\ {(3y)}^{3} - 3 \times {(3y)}^{2} \times \frac{1}{3x} + 3 \times 3y \times { (\frac{1}{3x}) }^{2} - {( \frac{1}{3x}) }^{3} = \\ 27 {y}^{3} - \frac{9 {y}^{2} }{x} + \frac{y}{ {x}^{2} } - \frac{1}{27 {x}^{3} }$