find the constant term of the expansion of (3x²+(1/x))^8

Question

I don't know where to start =\

Answer 1

(1/x^8) [ 3 x^10 +1 ]^8
now find the term in x^8 in the expansion of
(3 x^10 + 1)^8
because when you multiply by (1/x^8) it will be constant.
I would use the formula for binomial coefs rather than 9 rows of Pascal
C(n,r) = n!/[r!(n-r)!]
for n = 8 and r = 2
C = 8!/[6!(2!)]
= 8*7/2 = 28

Answer 2

The general term in the expansion is
t_r+1 = C(8,r)(3x²)^8-r(1/x)^r

= C(8,r) 3^(8-r) x^(16-3r)

to have a constant term the exponent of x must be zero, or
16-3r = 0

There is not integer solution for this, so the expansion does not have a "constant" term

another way to see it,...
the first term contains x^16, the second term constains x^13, .....
each term will contain an x term