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Kindly mention solution-techniques along with solution

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

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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 $4 = 8$? –  timvermeulen 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

1 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)

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How did you conclude that last bit? –  timvermeulen Jun 1 '13 at 19:54
    
It comes from the formula for a geometric sum. –  Cameron Williams Jun 1 '13 at 19:57
    
@timjver $1+x+\cdots+x^n=\dfrac{x^{n+1}-1}{x-1}$ –  Pedro Tamaroff Jun 1 '13 at 19:57
    
Please don't assume they will be in relations like x , x^2 , x^3 Then how to solve ? –  Arnab Dutta Jun 1 '13 at 20:15
    
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

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