Find the values of $a,b,c$ such that $(1+\sqrt[3]{4})/(2-\sqrt[3]{2})=a+b\sqrt[3]{2}+c\sqrt[3]{4}$

Question

Let $a,b,c\in\mathbb{Q}$. Find the values of $a,b,c$ such that $$(1+\sqrt[3]{4})/(2-\sqrt[3]{2})=a+b\sqrt[3]{2}+c\sqrt[3]{4}$$

What I tried:

I multiplied both sides by $(2-\sqrt[3]{2})$. I still have $3$ unknowns with only $1$ equation though.

Other thoughts:

If I get $0$ on one side of the equation, so it will end up looking (something along the lines of) $0=f(x)$. This is a constant but I can view it as if $a,b,c$ are variables. More importantly, I can think of them as roots of $f(x)$. Creating some sort of extension field of $\mathbb{Q}$ might be able to help me get the roots.

If this is the right way to approach the problem, I am not sure how to do this. If this is not the right way, I have absolutely no idea how to start. Any help would be appreciated. Thank you!

Answer 1

I still have $3$ unknowns with only $1$ equation, though.

Remember that: $$x+y\sqrt[3]2+z\sqrt[3]4=1+\sqrt[3]4$$ if and only if: $$x=1,y=0,z=1$$ (assuming $x,y,z\in\mathbb Q$).

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So, what did you get when you multiplied both sides by $2-\sqrt[3]2$, again? Correct me if I'm wrong, but you'd get: $$1+\sqrt[3]4=(2a-2c)+(2b-a)\sqrt[3]2+(2c-b)\sqrt[3]4$$ Do you know how to proceed from here?

Answer 2

Hint: $$(2-\sqrt[3]2)(2^2+2\sqrt[3]2 + \sqrt[3]4)=2^3-2=6$$

Answer 3

Recall that $$a^3-b^3 = (a-b)(a^2+ab+b^2)$$

If we let $a=2$ and $b=\sqrt[3]{2}$, then we can eliminate the cube root in $a-b$ by multiplying by $a^2+ab+b^2$.