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Let $P(x)=x^6+bx^5+cx^4+dx^3+ex^2+fx+g$ be a polynomial with integer coefficients.

$$P\left(\sqrt{2}+\sqrt[3]{3}\right)=0$$

$R(x)$ is the remainder of the division between $P(x)$ by $x^3-3x-1$.

Determine the sum of the coefficients of $R(x)$.

What i have tried/know $x=\sqrt 2 +\sqrt[3]3$ $\implies$

$(x-√2)^3=3$ $\implies$ $x^3-3x^2√2+6x-2√2=3$ $\implies$ $(x^3+6x-3)^2=√2^2(3x^2+2)^2$

Working that expression you get:

$x^6-6x^4-6x^3+12x^2-36x+1=0$

So the solution i saw said that this equation:

$x^6-6x^4-6x^3+12x^2-36x+1=P(x)$

Dividing by $x^3-3x-1.$ Writing as the remainder theorem:

$P(x)=(x^3-3x-5)*(x^3-3x-1)+(3x^2-54x-4)$ $R(1)= 3-54-4= -55$

Which is the answer

I dont get why can i grant P(x) is that polynomial. For me is just a polynomial with integer coefficients that happens to have $(√2+\sqrt[3]3)$ as a root. Is this granted to be unique?

I tried using the theorem of irrational roots (don't know if that's the name for it in English), but it only covers roots of the type $√a+√b$

Matthew Towers
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Rafael Guimarães
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Aug 13 '23 at 19:19

1 Answers1

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The minimal polynomial of $x=\sqrt2+\sqrt[3]3$, irrational of degree $6$ over $\mathbb Q$, is the result of simplify $(x-\sqrt2)^3=3$ so $$P(x)=x^6-6x^4-6x^3+12x^2-36x+1$$ and $$\frac{x^6-6x^4-6x^3+12x^2-36x+1}{x^3-3x-1}=x^3-3x+5+\frac{3x^2-54x-4}{x^3-3x-1}$$ Thus the required sum is equal to $3-54-4=-55$

Piquito
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  • How can i grant thats the polynomial P(x)? I get this is a polynomi tha share that root, but is there only one 6th degree polynomial with integer coefficients that has that root? What do you mean by minimal polynomial – Rafael Guimarães Aug 14 '23 at 12:37
  • @Rafael Guimarães This $P(x)=x^6-6x^4-6x^3+12x^2-36x+1$ is the only one of the sixth degree whose roots are the irrational $ \sqrt2+\sqrt[3]3$ and its conjugates. You have to study a little bit of algebraic number theory. Please don't argue that "another polynomial" could be $5P(x)$ or $47P(x)$. Regards. – Piquito Aug 14 '23 at 14:05
  • @Angelo It is a low-quality question for you and many other people including myself. But it is not for the OP. “Low-quality” is a relative thing in mathematics. Do you understand very clearly the Wiles-Taylor proof of the FLT? It could be that they see as low-quality what we do not consider as such. I like you as a member of MSE, your answers I have read I have appreciated. – Piquito Aug 14 '23 at 14:19
  • @Piquito, thank you very much for appreciating my answers, I also appreciate yours. When I wrote that OP’s post was a very low-quality question, OP had not written anything about what he had tried to solve his exercise, but, since now he has written his attempts and efforts, of course now his post is not a low-quality question any longer. Please, read the comment of “Another User”. – Angelo Aug 14 '23 at 14:38