Consider 1/(1+x)=(1+x)-1=1-x+x2-x3+x4-... Now integrate wrt x: x-x2/2+x3/3-x4/4+x5/5-...
But the integral of 1/(1+x) is ln(1+x), so ln(1+x) expands to x-x2/2+x3/3-x4/4+x5/5-...
Note that this formula is given in the instructions within the question; I've just showed where the formula came from.
ln(x+2)3=3ln(x+2)=3ln(2+x)=3ln[2(1+x/2)]=3ln(2)+3ln(1+x/2).
If we replace x by x/2 in the expansion of ln(1+x) we get:
x/2-x2/8+x3/24-x4/64+x5/160-...
From this we can deduce the expansion of ln(x+2)3=
3ln(2)+3x/2-3x2/8+x3/8-3x4/64+3x5/160-...
Expansion to the third term is 3ln(2)+3x/2-3x2/8.