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Given $P(x)=99(x^{101}-1)-101(x^{99}-1)$, find $Q(1)$ where $Q$ is the quotient of the division between $P(x)$ and $(x-1)^3$.

Obviously I could just divide the polynomials but that is not the solution I want. It is possible to figure out $Q(1)$ without doing long division, and that is the answer I am interested in.

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    $\begingroup$ I'm not quite clear on the question, actually. $P(x)=(x-1)^3Q(x)+R(x)$ where $R(x)$ is quadratic. To obtain $Q(1)$ differentiate each side three times with respect to $x$. $R$ vanishes and the derivatives of $Q$ get multiplied by zero when you put $x=1$. But is this what is intended? $\endgroup$ – Mark Bennet Aug 28 '17 at 21:18
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If we write: $$ P(x) = (x-1)^3Q(x)+ax^2+bx+c$$ then $$P'''(x) =6Q(x)+(x-1)[......]$$ On the other hand we have $$P'''(x)= 99\cdot 101 (100\cdot 99x^{98}-98\cdot 97x^{96})$$ So $$ 6Q(1) = P'''(1) = 99\cdot 101 (100\cdot 99-98\cdot 97)$$

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