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Решите задачу:

$\frac{c^2}{c^3-8}-\frac{5c+1}{8-c^3}-\frac{3-3c}{8-c^3}=\frac{c^2}{-(8-c^3)}-\frac{5c+1}{8-c^3}-\frac{3-3c}{8-c^3}=\frac{-c^2}{8-c^3}-\frac{5c+1}{8-c^3}-\frac{3-3c}{8-c^3}=\\\\=\frac{-c^2-(5c+1)-(3-3c)}{8-c^3}=\frac{-c^2-5c-1-3+3c}{8-c^3}=\frac{-c^2-2c-4}{2^3-c^3}=\frac{-(c^2+2c+4)}{2^3-c^3}=\\\\=\frac{-(c^2+2c+4)}{(2-c)\cdot(c^2+2c+2^2)}=\frac{-1}{2-c}=\frac{1}{-(2-c)}=\frac{1}{c-2}.$