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Решите задачу:

$\frac{x+1}{x}=a\; \; \to \; \; x+1=ax\; ,\; \; x(a-1)=1\; ,\; \; x= \frac{1}{a-1} \\\\ \frac{x^3+1}{x^3} = \frac{(x+1)(x^2-x+1)}{x^3} = \frac{ax\cdot (x^2-x+1)}{x^3} = \frac{a\cdot (x^2-x+1)}{x^2} =\\\\=\frac{a\cdot (\frac{1}{(a-1)^2}-\frac{1}{a-1}+1)}{(\frac{1}{a-1})^2} =\frac{a\cdot\frac{1-(a-1)+(a-1)^2}{(a-1)^2}}{\frac{1}{(a-1)^2}} =a\cdot (1-a+1+a^2-2a+1)=\\\\=a\cdot (a^2-3a+3)\ne a(a^2-3)\\\\ili\\\\ \frac{x^3+1}{x^3}= \frac{x^3}{x^3} + \frac{1}{x^3} =1+\frac{1}{\frac{1}{(a-1)^3}}=1+(a-1)^3=$

$=1+a^3-3a^2+3a-1=a^3-3a^2+3a=a\cdot (a^2-3a+3)\\\\a\cdot (a^2-3a+3)\ne a\cdot (a^2-3)$