A basic algebra question (no calculator)

AI Thread Summary
To simplify 3√-27x^6, the expression can be rewritten as (-27x^6)^(1/3). The cube root of -27 is -3, leading to the simplification of the expression to -3x^2. The confusion primarily revolved around calculating -27^(1/3) without a calculator. Understanding that -27 equals -3 cubed clarified the process for the participants. The discussion concluded with appreciation for the explanation provided.
Naga360
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Homework Statement



Simplify 3√-27x^6

I'm very confused on how to evaluate what -27^(1/3) is without a calculator. I think I'm fine with the other steps. if someone could explain this to me I would really appreciate it.



The Attempt at a Solution


= 3√-27x^6
= (-27x^6)^(1/3)
= -27^(1/3)x^2
 
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Naga360 said:

Homework Statement



Simplify 3√-27x^6

I'm very confused on how to evaluate what -27^(1/3) is without a calculator. I think I'm fine with the other steps. if someone could explain this to me I would really appreciate it.



The Attempt at a Solution


= 3√-27x^6
= (-27x^6)^(1/3)
= -27^(1/3)x^2

-27^\frac{1}{3} = -3^{3*\frac{1}{3}}
 
oh I get it now thanks a lot :)
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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