Solução:
Usa-se o teorema binomial. Em seguida, várias regras exponenciais para simplificar os termos.
( x 2 − 1 x ) 8 = ∑ i = 0 8 ( 8 i ) ( x 2 ) i ( − 1 x ) 8 − i {\displaystyle (x^{2}-{\frac {1}{x}})^{8}=\sum _{i=0}^{8}{\binom {8}{i}}(x^{2})^{i}({\frac {-1}{x}})^{8-i}} = ∑ i = 0 8 ( 8 i ) x 2 i ( − 1 ) 8 − i x 8 − i {\displaystyle =\sum _{i=0}^{8}{\binom {8}{i}}{\frac {x^{2i}(-1)^{8-i}}{x^{8-i}}}} = ∑ i = 0 8 ( 8 i ) x 3 i − 8 ( − 1 ) 8 − i {\displaystyle =\sum _{i=0}^{8}{\binom {8}{i}}x^{3i-8}(-1)^{8-i}} = x − 8 − 8 x − 5 + 28 x − 2 − 56 x 1 + 70 x 4 − 56 x 7 + 28 x 10 − 8 x 13 + x 16 {\displaystyle =x^{-8}-8x^{-5}+28x^{-2}-56x^{1}+70x^{4}-56x^{7}+28x^{10}-8x^{13}+x^{16}} = 1 x 8 − 8 x 5 + 28 x 2 − 56 x 1 + 70 x 4 − 56 x 7 + 28 x 10 − 8 x 13 + x 16 {\displaystyle ={\frac {1}{x^{8}}}-{\frac {8}{x^{5}}}+{\frac {28}{x^{2}}}-56x^{1}+70x^{4}-56x^{7}+28x^{10}-8x^{13}+x^{16}}
