Firstly sorry for this topic's title..
$${P(x)\over x^2}=x-1 \Rightarrow {P^3(x)\over x^2}=?$$
A question about fractional polynomials (two)
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algebra-precalculus
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5Hint: $P(x)=x^2(x-1)$. Therefore $(P(x))^3=?$. Therefore $\frac{(P(x))^3}{x^2}=?$. – 2011-11-13
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0P(x) same as p(x) ? – 2011-11-13
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0@daniel yes.... – 2011-11-13
1 Answers
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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 $$ {P^3(x)\over x^2}=x^4(x-1)^3. $$