Yes Jeff what you wrote is exactly what I intended to say, sorry I don't know how to make the cube root sign and everything else that you did. Posted below is the most up to date work that I have done to get the answer:
1 - \dfrac{1}{\sqrt[3]{r^2}}?[/tex] * cuberoot R^2*cuberootR^2= cuberootR^4=cuberootR^4-1(R/R)= cuberootR^4-R
_________________________________cuberootR^2*cuberootR^2 cuberootR^8=____R_________________R
I used two lines to indicate the numerator and the denominator. I hope this is makes sense. Please let me know if it doesn't. Thank you.
Please do NOT try to show fractions using the vertical notation unless you can use LaTeX. Otherwise, use the horizontal notation and PEMDAS. Unless you intend to ask many questions, it is probably not worth your while to learn LaTeX.
(a + b) / c = \(\displaystyle \dfrac{a + b}{c}\)
a + b/c = \(\displaystyle a + \dfrac{b}{c}.\)
As for roots, use ^ to indicate exponentiation and fractional exponents to represent roots: a^(1/3) = \(\displaystyle \sqrt[3]{a}.\)
I THINK that this is what you did.
\(\displaystyle 1 - \dfrac{1}{\sqrt[3]{r^2}} = 1 - \left(\dfrac{1}{\sqrt[3]{r^2}} * 1 * 1\right) = 1 - \left(\dfrac{1}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}}\right).\)
This first step is correct. When you multiply an expression by 1, you do not change the value of the expression.
After this you slip off the rails (if, that is, I am following you).
\(\displaystyle 1 - \left(\dfrac{1}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}}\right)\ DOES\ NOT\ EQUAL\ \dfrac{\sqrt[3]{r^4} - r}{r}.\)
\(\displaystyle 1 - \left(\dfrac{1}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^2}}{\sqrt[3]{r^2}}\right) = 1 - \left(\dfrac{1}{\sqrt[3]{r^2}} * \dfrac{\sqrt[3]{r^4}}{\sqrt[3]{r^4}}\right) = 1 - \dfrac{\sqrt[3]{r^4}}{\sqrt[3]{r^6}}= 1 - \dfrac{\sqrt[3]{r^4}}{r^2}.\)
There are still a few steps left to go. What do you get?
