Решите уравнение: (7a-6)/(a^3+27)=1/(a^2-3a+9)-1/(a+3).

\[\frac{7a - 6}{a^{3} + 27} = \frac{1}{a^{2} - 3a + 9} - \frac{1}{a + 3}\]

\[ОДЗ:\ \ x \neq - 3\]

\[\frac{7a - 6}{(a + 3)\left( a^{2} - 3a + 9 \right)} =\]

\[= \frac{a + 3 - \left( a^{2} - 3a + 9 \right)}{(a + 3)\left( a^{2} - 3a + 9 \right)}\]

\[7a - 6 = a + 3 - a^{2} + 3a - 9\]

\[7a - 6 = 4a - a^{2} - 6\]

\[a^{2} + 7a - 4a - 6 + 6 = 0\]

\[a^{2} + 3a = 0\]

\[a(a + 3) = 0\]

\[a = 0\ \ \ \ \ \ \ a + 3 = 0\]

\[Ответ:a = 0.\]