The question asks to find a Maclaurin series for the function
f(x) = (x-sinx)/x^3 when x does not = 0
f(x) = 1/6 when x = 0
Since sinx = x - x^3/3! + x^5/5! - ...
x-sinx = x^3/3! - x^5/5! + x^7/7! - ...
and (x-sinx)/x^3 = 1/3! - x^2/5! + x^4/7! - ... = 1/6 - x^2/5! + x^4/7! - ...
I noticed that I can make a series out of the second term onward, and in doing so would make f(x) = 1/6 when x = 0. This gives me 1/6 - series(n=0,infinity) ((-1)^(x+1) * x^(2n+2))/(2n+3)!
Would that be my answer? Just curious, because I've never done a problem like this before.