What if the degree of P was 3? Then P(x)=ax^3+bx^2+cx+d
Call z=5^(1/3)
so z^2=25^(1/3)=5^(2/3)
and z^3=5
x=z^2+z
x^2=z^2+5z+10
x^3=15z^2+15z+30
Expanding P(z^2+z)=(15a+b+c)z^2 + (15a+5b+c)z + (30a+10b+d)
This is a quadratic in z with integer coefficients. Can it equal the answers to a) or b)?
I'm not sure where to go from here...
I was trying so show P can't have degree 3.
For higher degree the same thing will happen: a quadratic in z with integer coefficients.
|
|
Posted by Jer
on 2021-05-17 14:52:41 |
A thought