Solve $(x-2)^6+(x-4)^6=64$ using a substitution. Solve $(x-2)^6+(x-4)^6=64$ using a substitution. 
I tired using $t=x-2$.
But again I have to know the expansion of the power $6$.
Can it be transformed to a quadratic equation ?
I know that it can be somehow solved by expanding the powers. But I'm trying to get a good transformation by a sub.

Let $x-2=a$ and $4-x=b$.
Hence, $a+b=2$ and $a^6+b^6=64$ or 
$$(a^2+b^2)(a^4+b^4-a^2b^2)=64$$ or
$$(2-ab)(a^2b^2-16ab+16)=32$$ or
$$ab(a^2b^2-18ab+48)=0.$$ 
$ab=0$ gives $x=2$ or $x=4$.
$ab=9-\sqrt{33}$ does not give a real roots and $ab=9+\sqrt{33}$ does not give a real roots.
Id est, the answer is $\{2,4\}$.
