\( (1+a^3)(a+1)^{-2} = \frac{1+a^3}{(a+1)^2} = \frac{(1+a)(1-a+a^2)}{(a+1)^2} = \frac{1-a+a^2}{a+1} \)
\( (x^2-y^2):(x^{-1}-y^{-1}) = (x^2-y^2) : (\frac{1}{x} - \frac{1}{y}) = (x-y)(x+y) : (\frac{y-x}{xy}) = (x-y)(x+y) \cdot \frac{xy}{y-x} = -(x-y)(x+y) \cdot \frac{xy}{x-y} = -(x+y)xy = -xy(x+y) \)
\( (\frac{a}{c})^{-1} + (\frac{c}{a})^{-1} = \frac{c}{a} + \frac{a}{c} = \frac{c^2+a^2}{ac} \)
\( (\frac{1}{b^3} + \frac{1}{c^3})(b+c)^{-1} = (\frac{c^3+b^3}{b^3c^3}) \cdot \frac{1}{b+c} = \frac{(c+b)(c^2-cb+b^2)}{b^3c^3(b+c)} = \frac{c^2-cb+b^2}{b^3c^3} \)
Ответ: 1) а) \( \frac{1-a+a^2}{a+1} \), б) \( -xy(x+y) \); 2) а) \( \frac{c^2+a^2}{ac} \), б) \( \frac{c^2-cb+b^2}{b^3c^3} \).