Question

27a^3-b^3/9a^2+3ab+b^2

Answer 1

Объяснение:

$\frac{27a {}^{3} - b {}^{3} }{9a {}^{2} + 3ab + b {}^{2} } = \frac{(3a - b)(9a {}^{2} + 3ab + b {}^{2} )}{9a {}^{2} + 3ab + b {}^{2} } = 3a - b$

Answer 2

$\displaystyle\bf\\\frac{27a^{3} -b^{3} }{9a^{2} +3ab+b^{2} } =\frac{(3a)^{3} -b^{3} }{9a^{2} +3ab+b^{2} } =\frac{(3a-b)(9a^{2} +3ab+b^{2}) }{9a^{2} +3ab+b^{2} } =\\\\\\=3a-b$

При решении была применена формула разности кубов :

$\dfisplaystyle\bf\\a^{3} -b^{3} =(a-b)(a^{2} +ab+b^{2} )$