Solve this system of equations...

Question

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

Answer 1

$$3pa^2+3qb^2=3\Rightarrow pa^2+qb^2=1$$ 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.

Answer 2

$(p,q)$ 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.