"Spoooooooky" Russian rational expressions 31

Algebra Level 3

$\large \dfrac{8-N^3}{N+2} \div \left( 2 + \dfrac{N^2}{N+2} \right) - \dfrac{N^2}{N-2}\cdot \dfrac{4-N^2}{N^2 + 2N} = \, ?$

Evaluate the above when $N = 2016$ .

1 solution

Arkajyoti Banerjee
Jun 28, 2016

Let $n=2016$ ,

Moving forward to the solution by abiding by the VBODMAS rule, we have:

$\displaystyle \frac{8-n^{3}}{2+n} \times \frac{2+n}{n^{2}+2n+4} + \frac{n^{2}}{2-n} \times \frac{4-n^{2}}{n^{2}+2n}$

$\displaystyle =(2-n)(n^{2}+2n+4) \times \frac{1}{n^{2}+2n+4} + \frac{n^{2}}{2-n} \times \frac{(n+2)(2-n)}{n(n+2)}$

Simplifying and cancelling terms out, we have:

$\displaystyle =(2-n)+(n)=\boxed{2}$

It requires the value of $n \neq 0, 2 , -2$ lol :)

Calvin Lin Staff - 4 years, 11 months ago

Oh yeah! :D

Arkajyoti Banerjee - 4 years, 11 months ago

Ya did in the same way.....................

Abhisek Mohanty - 4 years, 11 months ago

I just put the value of n as 2016 because I didn't want to hint at anything about the problem.

Hobart Pao - 4 years, 11 months ago

And I didn't even use the fact that $N=2016$ Ha.

Bloons Qoth - 4 years, 11 months ago