# Simplify

• Jun 2nd 2012, 08:33 PM
zbest1966
Simplify
(c+1/c2)2

= (c+1/c)(c+1/c)

= c2+1/c2+1/c2+1/c4

=c2+2/c2+1/c4

=c6+2c2+1
• Jun 2nd 2012, 09:33 PM
rainbowoblivion
Re: Simplify
Are you just wanting someone to double check your work?
• Jun 2nd 2012, 10:00 PM
Prove It
Re: Simplify
Quote:

Originally Posted by zbest1966
(c+1/c2)2

= (c+1/c)(c+1/c)

= c2+1/c2+1/c2+1/c4

=c2+2/c2+1/c4

=c6+2c2+1

I suggest you double check your question. Are you trying to expand \displaystyle \displaystyle \begin{align*} \left(c + \frac{1}{c}\right)^2 \end{align*} or \displaystyle \displaystyle \begin{align*} \left(c + \frac{1}{c^2}\right)^2 \end{align*}?
• Jun 3rd 2012, 06:58 AM
Soroban
Re: Simplify
Hello, zbest1966!

Wow! . . . Can you make a few more errors ... please?

Quote:

$\displaystyle \left(c+\frac{1}{c^2}\right)^2$

. . $\displaystyle =\;\left(c+\frac{1}{c}\right)\left(c+\frac{1}{c} \right)$ . Where is the $\displaystyle {\color{blue}c^2}$?

. . $\displaystyle =\; c^2 + \frac{1}{c^2}+\frac{1}{c^2} +\frac{1}{c^4}$ . This is terribly wrong!

. . $\displaystyle =\; c^2 + \frac{2}{c^2} +\frac{1}{c^4}$ . Well, you added correctly ...

. . $\displaystyle =\; c^6 + 2c^2 +1$ . What happened to the denominator?

I will assume that the problem is: .$\displaystyle \left(c + \frac{1}{c^2}\right)^2$

We have: .$\displaystyle \left(c + \frac{1}{c^2}\right)\left(c + \frac{1}{c^2}\right)$

"FOIL": .$\displaystyle (c)(c) + (c)\left(\frac{1}{c^2}\right) + \left(\frac{1}{c^2}\right)(c) + \left(\frac{1}{c^2}\right)\left(\frac{1}{c^2} \right)$

. . . . . $\displaystyle =\;c^2 + \frac{1}{c} + \frac{1}{c} + \frac{1}{c^4}$

. . . . . $\displaystyle =\;c^2 + \frac{2}{c} + \frac{1}{c^4}$

. . . . . $\displaystyle =\;\frac{c^6 + 2c^3 + 1}{c^4}$
• Jun 3rd 2012, 07:01 AM
Wilmer
Re: Simplify