Let $P(x^{3})=ax^{6}+(b-2)x^{5}+(a+3)x^{4}+bx^{3}+3$ be a polynomial.Then, what is the remainder when $P(x)$ is divided by $(x-3)$

Question

I have run into a question in my textbook, but I could not solve it. I tried classical methods but i could not reach the answer. I think that the question is deficit from enough information. Can you give me hints or solution.The question is:

Let $P$ be a polynomial over $\mathbb{R}$ such that $$P(x^{3})=ax^{6}+(b-2)x^{5}+(a+3)x^{4}+bx^{3}+3,$$ for some $a,b\in \mathbb{R}$.

Then, what is the remainder when $P(x)$ is divided by $(x-3)$

Note: The answer is $-18.$

Answer 1

If $P(x)$ is a polynomial, $P(x^3)$ will have only terms where the exponent of $x$ is a multiple of $3$. Note this is slightly different from how you have phrased your question.

Hence $b=2, a=-3$ and $P(x^3) = -3x^6+2x^3+3 \implies P(3) = -3\cdot(3)^2+2\cdot(3)+3=-18.$