Show this equation has no rational solution $ (x^{39}+7)^3-2(x^{39}+6)+1=0$ I know that i can use the rational root theorem, and substitution, but I don't know how to connect with this equation.  Can someone help me with that?
 A: Let $y=x^{39}+7$. Then you have $$y^3-2(y-1)+1=y^3-2y+3=0$$ This has no rational root, since if it did, the rational root theorem would require a root to be $\pm1$ or $\pm3$. None qualifies.
So $y=x^{39}+7$ is not rational. So $x^{39}$ is not rational. So $x$ is not rational.

Note that since we have a quadratic-suppressed cubic with only one real root, it would actually be not so hard to solve for $y$ exactly using the cubic formula. However this wouldn't directly tell you that $y$ is not rational.
A: First, let's expand that polynomial to make things easier:
$(x^{39}+7)^3-2(x^{39}+6)+1=0$
Becomes
$x^{117}+21 x^{78}+145 x^{39}+332 = 0$
The rational root theorem means that all of the rational solutions to your equation have the form $\pm p/q$, where $p$ is an integer factor of 332 (the coefficient of $x^0$) and $q$ is a factor 1 (the coefficient of $x^{117}$. The factors of 332 are: 1, 2, 4, 83, 166, and 332. So, the possible rational roots are:

$1, -1, 2, -2, 4, -4, 83, -83, 166, -166, 332, -332$

At this point, it's trivial to prove that none of those is a solution. But those were all of the possible rational solutions. Thus, there is no rational solution.

A better approach: 
One might object that calculating $332^{117}$ is a bit hairy. True enough. This is where substitution is handy. Let $y = x^{39}$. Then the equation is:
$y^{3}+21 y^{2}+145 y+332 = 0$
The same logic as above gives us the same possible solutions, only this time they are possible values for $y$. Again, none of them works, which means there is no rational root for this equation in terms of $y$. But $y = x^{39}$ is rational for all rational $x$. Thus, if no rational $y$ is a root, then there also cannot be a rational $x$ that satisfies the original equation. 
