Simplify

**zbest1966** · June 2nd 2012, 08:33 PM

(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

**rainbowoblivion** · June 2nd 2012, 09:33 PM

Are you just wanting someone to double check your work?

**Prove It** · June 2nd 2012, 10:00 PM

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 $\displaystyle \begin{align*} \left(c + \frac{1}{c}\right)^2 \end{align*}$ or $\displaystyle \begin{align*} \left(c + \frac{1}{c^2}\right)^2 \end{align*}$ ?

**Soroban** · June 3rd 2012, 06:58 AM

Hello, zbest1966!

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

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

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

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

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

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

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

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

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

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

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

. . . . . $=\;\frac{c^6 + 2c^3 + 1}{c^4}$

**Wilmer** · June 3rd 2012, 07:01 AM

...or [(c + 1) / c^2]^2 ?

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