Going by the Irrational Root Theorem:
Let a and b be rational numbers and let √b be an irrational number.
If a + √v=b is a root of a polynomial equation with rational coefficients then the conjugate a - √b also is a root.
Here's my question:
what if its 3^√3? The conjugate would be -3^√3 would it not? but when put into the equation x^3 - 3 = 0. 3^√3 works but not its conjugate. I don't understand why.
Jhevon showed you why it doesn't work. I may add that doesn't fit your theorem. Would that be ? or ? But even in these case, where is the other term? You cannot apply the theorem in such an example. That's why it doesn't have to work! And it doesn't. Never heard of that theorem, but seems nice.Going by the Irrational Root Theorem:
Let a and b be rational numbers and let √b be an irrational number.
If a + √v=b is a root of a polynomial equation with rational coefficients then the conjugate a - √b also is a root.
Here's my question:
what if its 3^√3? The conjugate would be -3^√3 would it not? but when put into the equation x^3 - 3 = 0. 3^√3 works but not its conjugate. I don't understand why.
It is actually the irrational conjugate roots theorem