TITLE: Rational number trivial case
QUESTION [5 upvotes]: Let $a,b,c$ denote rational numbers, such that $(a+b\sqrt[3]2+c\sqrt[3]4)^3$ is also rational. Prove that at least two of the numbers $a,b,c$ must be zero. Actually I confused of the beginning steps for this proof. Should I apply the strategies of contradictory proofs

REPLY [2 votes]: Oh an answer while I was writing! (I was writing in Notepad not directly in the forum).
Anyway, before answering the question, lets prove these:
Lemma 1: $\sqrt[3]{2}\not\in\mathbb{Q}^+$.
Corollary: $\sqrt[3]{4}\not\in\mathbb{Q}^+$.
Lemma 2: $\forall (a,b)\in\mathbb{Z}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longleftrightarrow (a,b)=(0,0)$
Theorem: $\forall (a,b)\in\mathbb{Q}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longleftrightarrow (a,b)=(0,0)$
Proof of lemma 1:
Assume that $\sqrt[3]{2}\in\mathbb{Q}$. Then $\exists !(a,b)\in\mathbb{N}^2,\,\sqrt[3]{2}=\dfrac{a}{b}$ such as $a,b\neq 0$ and $a∧b=1$.
Then $$a^3=2b^3\,\,\,\,(1)$$and so $a$ is an even number. Then $\exists k\in\mathbb{N},\,a=2k$ and so in $(1)$ we have $b^3=4k^3$ and so $b$ is also an even number which contradicts $a∧b=1$.
Proof of the corollary:
Assume that $\sqrt[3]{4}\in\mathbb{Q}^+$. Then $\sqrt[3]{4^2}\in\mathbb{Q}^+$ and so $2\sqrt[3]{2}\in\mathbb{Q}^+$ which contradicts the lemma 1.
Proof of lemma 2:
If $(a,b)=(0,0)$ then $a\sqrt[3]{2}+b\sqrt[3]{4}=0\in\mathbb{Q}$.
Now lets prove that $\forall (a,b)\in\mathbb{Z}^2,\,a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\Longrightarrow (a,b)=(0,0)$.
To prove it, lets prove that $\forall (a,b)\in\mathbb{Z}^2,\,(a,b)\neq (0,0)\Longrightarrow a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$
If $a=0$, and so $b\neq 0$, or $b=0$ and so $a\neq 0$, it's trivial that $a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$ using the lemma 1 or the corollary.
Assume that $\exists (a,b)\in\mathbb{Z}^2$ such as $a,b\neq 0$ and: $$a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}\,\,\,\, (2.1)$$
Then $(a\sqrt[3]{2}+b\sqrt[3]{4})^2\in\mathbb{Q}$ and $(a\sqrt[3]{2}+b\sqrt[3]{4})^2=a^2\sqrt[3]{4}+2b^2\sqrt[3]{2}+4ab$.
While $4ab\in\mathbb{N}$ then: $$a^2\sqrt[3]{4}+2b^2\sqrt[3]{2}\in\mathbb{Q}\,\,\,\, (2.2)$$
By $(2.2)-\dfrac{a^2}{b}(2.1)\in\mathbb{Q}\,\,\,\,\left(\dfrac{a^2}{b}\in\mathbb{Q}^+\right)$ we get: $$\left( 2b^2-\dfrac{a^3}{b}\right)\sqrt[3]{2}\in\mathbb{Q}\,\,\,\, (2.3)$$
If $\left( 2b^2-\dfrac{a^3}{b}\right) =0$ then $a=\sqrt[3]{2}b\not\in\mathbb{N}$ which is false.
And so $\left( 2b^2-\dfrac{a^3}{b}\right)\neq 0$, but while $\left( 2b^2-\dfrac{a^3}{b}\right)\in\mathbb{Q}$ it means using $(2.3)$ that $\sqrt[3]{2}\in\mathbb{Q}$ which contradicts the lemma 1.
We conclude that $\forall (a,b)\in\mathbb{Z}^2,\,(a,b)\neq (0,0)\Longrightarrow a\sqrt[3]{2}+b\sqrt[3]{4}\not\in\mathbb{Q}$ which proves lemma 2.
Proof of the theorem:
If $(a,b)=(0,0)$ it's as same as in lemma 2.
Let $(a,b)\in\mathbb{Q}$ such as $a\sqrt[3]{2}+b\sqrt[3]{4}\in\mathbb{Q}$.
$(a,b)\in\mathbb{Q}$ so $\exists ((p,p'),(q,q'))\in\mathbb{Z}^2\times\mathbb{N}^{+2}\, (a,b)=\left(\dfrac{p}{q},\dfrac{p'}{q'}\right)$.
We have then:$$\dfrac{p}{q}\sqrt[3]{2}+\dfrac{p'}{q'}\sqrt[3]{4}\in\mathbb{Q}\,\,\,\, (3)$$
By $qq'\times (3)$ still $qq'\in\mathbb{N}^+$ we get $pq'\sqrt[3]{2}+p'q\sqrt[3]{4}\in\mathbb{Q}$. Using lemma 2 we get $(pq',p'q)=(0,0)$ and so $(p,p')=(0,0)$ and so $(a,b)=(0,0)$ which proves the theorem.
We know that: $$\forall (x,y,z)\in\mathbb{R}^3,\,(x+y+z)^3=x^3+y^3+z^3+3x^2(y+z)+3y^2(x+z)+3z^2(x+y)+6xyz\,(4)$$.
Now let $(a,b,c)\in\mathbb{Q}^3$ such as $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$.
By using $(4)$: $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3=a^3+2b^3+4c^3+12abc+3(a^2b+2b^2c+2ac^2)\sqrt[3]{2}+3(a^2c+ab^2+2bc^2)\sqrt[3]{4}$.
While $a^3+2b^3+4c^3+12abc\in\mathbb{Q}$, $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$ if and only if $3(a^2b+2b^2c+2ac^2)\sqrt[3]{2}+3(a^2c+ab^2+2bc^2)\sqrt[3]{4}\in\mathbb{Q}$.
Using the theorem: $(a+b\sqrt[3]{2}+C\sqrt[3]{4})^3\in\mathbb{Q}$ if and only if $a^2c+ab^2+2bc^2=0$ and $a^2b+2b^2c+2ac^2=0$.
Then see Jack D'Aurizio's post.