- #1

- 274

- 1

## Homework Statement

I have the equation

$$

a^3+2b^3+4c^3-6abc=0

$$

and I want to show that for ##a,b,c\in\mathbb{Q}## and ##a,b,c\not=0## that there are no solutions.

## The Attempt at a Solution

I tried many different approaches but this was the only one that seemed to work. I am not sure if this is even allowed.

Holding ##a,c## constant, I first differentiate the equation in terms of ##b##. Then I have

$$

6b^2=6ac\Leftrightarrow b=\sqrt{ac}.

$$

Now I differentiate the original equation in terms of ##a## holding ##b,c## constant so

$$

3a^2=6bc\Leftrightarrow a^2=2bc\Leftrightarrow a=\sqrt{2bc}\Leftrightarrow a=\sqrt{2c\sqrt{ac}}\Leftrightarrow a^4=4c^3a\Leftrightarrow a^3=4c^3.

$$

Plugging this back into the original equation gives me

$$

a^3+2b^3+4c^3=2b^3+8c^3=6abc.

$$

Differentiating again in terms of ##b## holding ##c## constant gives me

$$

b^2=ac\Leftrightarrow b^2=c\sqrt{2bc}\Leftrightarrow b^4=2bc^3\Leftrightarrow b^3=2c^3.

$$

Plugging this back in gives

$$

2b^3+8c^3=12c^3=6abc

$$

so

$$

12c^3=6abc\rightarrow 2c^2=ab\Leftrightarrow 2c^2=8c^6\Leftrightarrow c^4=\frac{1}{4}\Leftrightarrow c=\frac{1}{\sqrt[4]{4}}.

$$

Now ##\sqrt[4]{4}## is irrational so ##\frac{1}{\sqrt[4]{4}}## is irrational and hence if ##a,b,c\not=0## there are no solutions.

I might also have to give more justification to the ##\sqrt[4]{4}## part as well but is my main idea ok? Am I allowed to differentiate like that?