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Find numbers $a$ , $b$ , $p$ and $q$ so that: $$x^3+15x^2+3x+5=p(x-a)^3+q(x-b)^3$$

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  • $\begingroup$ It would be useful if you could tell us what you have tried. This helps us to understand what you already know so we can better focus our help. Also, people around here get touchy when people just post a question without motivation etc. as they are unwilling to simply do others homework for them, moreover they dislike it when a third party steps in and does the homework for them anyway so questions like this which merely states a question tend to get closed (pending adding more info). Adding in what you have tried and where you came across it will stop your question getting closed in this way. $\endgroup$
    – user1729
    Oct 26, 2013 at 12:49
  • $\begingroup$ With a few steps I got: $$p(x-a)^3+q(x-b)^3=x^3(p+q)+x^2(-3ap-3bq)+x(3pa^2+3qb^2)-pa^3-qb^3$$Then I wrote a system of 4 equations, but didn't know how to solve them. $\endgroup$
    – A6SE
    Oct 26, 2013 at 12:55
  • $\begingroup$ Hint: this equation should hold for all $x$, in particular, for $0,\,1\,-1$. In addition, $p+q=1$. $\endgroup$ Dec 18, 2013 at 12:10

2 Answers 2

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$$3pa^2+3qb^2=3\Rightarrow pa^2+qb^2=1$$ You have also $p+q=1$. So $pa^2+qb^2=p+q=1$ from here we conclude that $a=\pm1,b=\pm1$. $$-3pa-3qb=15\Rightarrow pa+qb=-5$$

Using $pa+qb=-5$ and $p+q=1$ with $a=1,b=-1$ you find $p=-2$ and $q=+3$

Using $pa+qb=-5$ and $p+q=1$ with $a=-1,b=1$ you find $p=3$ and $q=-2$. Therefore the solution is $$(a,b,p,q)=(1,-1,-2,3) \quad \& \quad(a,b,p,q)=(-1,1,3,-2)$$

This answer is for the question before edited.

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$(p,q)$ and $(pa^2,qb^2)$ both obey the same pair of linear equations.
So either the pair of equations is degenerate, or the two vectors are equal.

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  • $\begingroup$ yeah, it seems like that. I'll edit my first post to the original task and pls see if you can help me then... $\endgroup$
    – A6SE
    Oct 26, 2013 at 12:44
  • $\begingroup$ No, I meant that $(p,q)$ obeys equations 1 and 2, and $(pa^2,qb^2)$ obeys equations 3 and 4. $\endgroup$
    – Empy2
    Oct 26, 2013 at 12:47

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