(x - 4)^5 - 9(x - 4)^3 = 0
find all real solutions for x
To visualize this better, let $\displaystyle u = x - 4$.
So our equation becomes: $\displaystyle u^5 - 9u^3 = 0$
Factor out $\displaystyle u^3$ to get: $\displaystyle u^3 (u^2 - 9) = 0$
And notice that the second factor is a difference of squares. Can you carry on from here?