$\displaystyle
pi^2/-s^3-2 = - (pi^2/-s^3-2)
$
$\displaystyle
= pi^2/ -s^3+2
$
am i using the subtraction correctly?
Hello,
I don't know what you mean by /- but I assume you left out some parentheses.
pi^2/(-s^3-2)=$\displaystyle \frac{pi^2}{-s^3-2}=-\frac{pi^2}{s^3+2}$.
pi^2/(-s^3)-2=$\displaystyle \frac{pi^2}{-s^3}-2=-\frac{pi^2}{s^3}-2$.
-(pi^2/(-s^3-2))=$\displaystyle -\frac{pi^2}{-s^3-2}=\frac{pi^2}{s^3+2}$.
-(pi^2/(-s^3)-2))=$\displaystyle -\left(\frac{pi^2}{-s^3}-2\right)=\frac{pi^2}{s^3}+2$.
You wrote pi^2/-s^3-2= pi^2/ -s^3+2 which seems obviously wrong.
Bye.
Hello,
$\displaystyle -(A+B)=-A-B, -(A-B)=-A+B,-(-A-B)=A+B$,
$\displaystyle -\frac{A}{B}=\frac{A}{-B}=\frac{-A}{B}$.
Thus,
$\displaystyle -\frac{A}{B+C}=\frac{A}{-(B+C)}=\frac{A}{-B-C}$,
$\displaystyle -\frac{A}{B-C}=\frac{A}{-(B-C)}=\frac{A}{-B+C}$,
$\displaystyle -\frac{A}{-B-C}=\frac{A}{-(-B-C)}=\frac{A}{B+C}$.
I hope it helps. I wrote all the possibilities in the previous post.
Bye.
If you saying the left hand side equals the bottom right hand side then nope.
You can see this is not true by plugging in a value for s. Try plugging in 1 and you will notice
$\displaystyle \frac{\pi^2}{-(1)^3 - 2}~\text{does not equal}~ \frac{\pi^2}{-(1)^3+2}$
Testing to see if it even
$\displaystyle y = \frac{\pi^2}{-s^3-2}$
Replace s with -s
$\displaystyle y = \frac{\pi^2}{-(-s)^3-2}$
which equals
$\displaystyle y = \frac{\pi^2}{s^3-2}$
so it is not the same as the original equation so it is not even.
To see if it odd replace y with -y and s with -s
$\displaystyle -y = \frac{\pi^2}{-(-s)^3-2}$
$\displaystyle -y = \frac{\pi^2}{s^3-2}$
$\displaystyle y = \frac{\pi^2}{-s^3+2}$
which is not the same as the original function so it is neither
Ok sorry I misread
$\displaystyle y = \frac{\pi^2}{s^3-2}$
To see if it odd replace y with -y and s with -s
$\displaystyle -y = \frac{\pi^2}{(-s)^3-2}$
$\displaystyle -y = \frac{\pi^2}{-s^3-2}$
$\displaystyle y = \frac{\pi^2}{s^3+2}$
So nope it is not odd either