Question

help!?!?!

Answer 1

the numerator is a difference of squares.

Answer 2

\[\large \frac{1}{36}-\frac{1}{x^2}=\left(\frac{1}{6}-\frac{1}{x}\right)
\left(\frac{1}{6}+\frac{1}{x}\right) \]

Answer 3

and then you can simplify those terms
\[\frac{ \left(\frac{1}{6}-\frac{1}{x}\right) \left(\frac{1}{6}+\frac{1}{x}\right) }{\left(\frac{1}{6}+\frac{1}{x}\right) }\]1/6 + 1/x cancels up and down.

we're left with 1/6 - 1/x and then finally find a mutual denominator. the mutual denominator is the GCF of 6 and x. to achieve result extend 1/6 with x/x and extend 1/x with 6/6 to get a mutual denominator

Answer 4

c