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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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    @daniel yes....2011-11-13

1 Answers 1

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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. $