Rational roots theorem to prove irrational

[srfriggen]

Homework Statement

Use the rational roots theorem to prove 3^1/2-2^1/3is irrational.

The Attempt at a Solution

My teacher strongly hinted to us that this problem had something to do with the fact that complex roots come in conjugate pairs, and all we had to do was, "flip the sign". But there isn't anything complex about this.

I was trying to work it out by setting x=3^1/2-2^1/3and trying to get rid of the fractional exponents by squaring or cubing, but I can't get the algebra down to get that to work.

 

[FeDeX_LaTeX]

Your method does work -- try cubing to get rid of the cube root first, then squaring to deal with the square root. The result then follows from the rational root theorem (your constant term will be prime and your leading co-efficient a one, giving you only four cases to verify).

 

[srfriggen]

Your method does work -- try cubing to get rid of the cube root first, then squaring to deal with the square root. The result then follows from the rational root theorem (your constant term will be prime and your leading co-efficient a one, giving you only four cases to verify).

I'm getting a TON of terms when I do it this way, am I doing something wrong, or should I just be extremely careful with the algebra and expect it to be messy?

 

[FeDeX_LaTeX]

Expand $(\sqrt{3} - x)^3 = 2$ (there's your cubing of both sides). The LHS should give you some square roots which you can put together on one side and square to get rid of them. The result should be a polynomial of order 6.