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Firstly sorry for this topic's title..
$${P(x)\over x^2}=x-1 \Rightarrow {P^3(x)\over x^2}=?$$

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Hint: $P(x)=x^2(x-1)$. Therefore $(P(x))^3=?$. Therefore $\frac{(P(x))^3}{x^2}=?$. – André Nicolas Nov 13 '11 at 19:18
P(x) same as p(x) ? – daniel Nov 13 '11 at 19:21
@daniel yes.... – Yusuf Ali Bozkır Nov 13 '11 at 19:28

Cubing both sides of your first equation yields $${P^3(x)\over x^6}=(x-1)^3.$$ Multiplying both sides of the above by $x^4$ gives $$ {P^3(x)\over x^2}=x^4(x-1)^3. $$

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