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

Kindly mention solution-techniques along with solution

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## marked as duplicate by Arthur Fischer♦Jun 3 '13 at 13:28

And your contribution is ...? – L. F. Jun 1 '13 at 19:42
Hint: $4=2^2$, $8=2^3$, power laws. – celtschk Jun 1 '13 at 19:46
@celtschk So that means that $4 = 8$? – Tim Vermeulen Jun 1 '13 at 19:50
@timjver: Oops, that was a typo, now fixed, thanks ;-) – celtschk Jun 1 '13 at 19:51
removed comment – Arnab Dutta Jun 1 '13 at 20:06

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)
@timjver $1+x+\cdots+x^n=\dfrac{x^{n+1}-1}{x-1}$ – Pedro Tamaroff Jun 1 '13 at 19:57
Arnab Dutta:$(\frac{1}{2})^p+(\frac{1}{2})^{p^2}+(\frac{1}{2})^{p^3}=1 $$\to$$\frac{({\frac{1}{2}})^{p^4}-1}{({\frac{1}{2}})^p-1}=2$ then use discriminant – Maisam Hedyelloo Jun 1 '13 at 20:22