Hi what is (3x + 1/x)^2 ???????
Hi!
In addition to Wilmer's post
One should lnow the binimial theorems
$\displaystyle (a+b)^2 = a^2 + 2ab + b^2$
$\displaystyle (a-b)^2 = a^2 - 2ab + b^2$
$\displaystyle (a+b)(a-b) = a^2-b^2$
In the example it is a = 3x and b = 1/x
Can you solve it now?
Edit
Okay, here is the solution
(3x)^2+2*(3x)*(1/x)+(1/x)^2
$\displaystyle = 9x^2+\frac{6x}{x} + \frac{1^2}{x^2}$
$\displaystyle = 9x^2 +6 + \frac{1}{x^2}$