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

I don't know where to start =\

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  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

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  2. The general term in the expansion is
    tr+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

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