(c+1/c^{2})2
= (c+1/c)(c+1/c)
= c^{2}+1/c^{2}+1/c^{2}+1/c^{4}
=c^{2}+2/c^{2}+1/c^{4 }=c^{6}+2c^{2}+1
Hello, zbest1966!
Wow! . . . Can you make a few more errors ... please?
$\displaystyle \left(c+\frac{1}{c^2}\right)^2$
. . $\displaystyle =\;\left(c+\frac{1}{c}\right)\left(c+\frac{1}{c} \right)$ . 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": .$\displaystyle (c)(c) + (c)\left(\frac{1}{c^2}\right) + \left(\frac{1}{c^2}\right)(c) + \left(\frac{1}{c^2}\right)\left(\frac{1}{c^2} \right)$
. . . . . $\displaystyle =\;c^2 + \frac{1}{c} + \frac{1}{c} + \frac{1}{c^4}$
. . . . . $\displaystyle =\;c^2 + \frac{2}{c} + \frac{1}{c^4}$
. . . . . $\displaystyle =\;\frac{c^6 + 2c^3 + 1}{c^4}$
...or [(c + 1) / c^2]^2 ?
Video: How to Use Parentheses in Equations | eHow.com