Is ##9\sqrt[3]{-3}## Equivalent to ##-9\sqrt[3]{3}##?

  • Thread starter Thread starter RChristenk
  • Start date Start date
AI Thread Summary
The discussion centers on whether ##9\sqrt[3]{-3}## is equivalent to ##-9\sqrt[3]{3}##. It is clarified that when taking the cube root of a negative number, the result is negative, allowing the negative sign to be moved outside the radical as a simplification. A reference example is provided, showing that ##\sqrt[3]{-8} = -2##, confirming the principle. The usual convention is to express the negative sign outside the radical for clarity. Thus, the two expressions are indeed equivalent.
RChristenk
Messages
73
Reaction score
9
Homework Statement
Find ##\sqrt[3]{-2187}##
Relevant Equations
None
I calculated this to be ##9\sqrt[3]{-3}##, but the answer is given as ##-9\sqrt[3]{3}##. Are these two quantities equal? If so, what is the usual convention for placement of the negative sign? Thanks.
 
Physics news on Phys.org
When you take a cube root or other odd root of -1, you get -1 (at least until you get to the chapter on complex numbers). So, in those cases, moving the sign outside the radical is considered a simplification.
 
  • Like
Likes PeroK and RChristenk
Just FYI, I changed the title from square root to cube root. :wink:
 
RChristenk said:
Homework Statement: Find ##\sqrt[3]{-2187}##
Relevant Equations: None

I calculated this to be ##9\sqrt[3]{-3}##, but the answer is given as ##-9\sqrt[3]{3}##. Are these two quantities equal? If so, what is the usual convention for placement of the negative sign? Thanks.
You could look at a graph of the cube root function:

https://www.cuemath.com/calculus/cube-root-function/

And you'll see that the cube root of a negative number is just a regular negative number.
 
A simpler example is this: ##\sqrt[3]{-8} = -2##. As a check, cube the result on the right side.
##(-2)^3 = (-2)(-2)(-2) = -8##
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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...
Back
Top