How to find $p$ when $ ({\frac{1}{2}})^p + ({\frac{1}{4}})^p + ({\frac{1}{8}})^p - 1 = 0. $

Question

Kindly mention solution-techniques along with solution

Answer 1

Hint: if $({\frac{1}{2}})^p + ({\frac{1}{4}})^p + ({\frac{1}{8}})^p = 1 $, let $ x=({\frac{1}{2}})^p$, then $$x+x^2+x^3=1\to \frac{x^4-1}{x-1}=2$$ we can find all Roots of a cubic function by using the discriminant (see here http://en.wikipedia.org/wiki/Cubic_function)