Determine the constant term in the following expression? binomial theorem

(x + 1/x + 1/x^2)^8

the answer is 238, but how do you get it?

Note that:
(x + 1/x + 1/x^2)^8 = [(x^3 + x + 1)/x^2]^8
= (x^3 + x + 1)^8/x^16.

This shows that the constant term of (x + 1/x + 1/x^2)^8 is the coefficient of the x^16 term in (x^3 + x + 1)^8.

By applying the Binomial Theorem twice
(x^3 + x + 1)^8 = [(x^3 + x) + 1]^8
= ∑ [C(8, k) * (x^3 + x)^(8 - k) * 1^k] (from k=0 to 8)
= ∑ {C(8, k) * ∑ [C(8 - k, l) * (x^3)^(8 - k - l) * x^k] (from l=0 to 8)} (from k=0 to 8)
= ∑ {C(8, k) * ∑ [C(8 - k, l) * x^(24 - 2k - 3l)] (from l=0 to 8)} (from k=0 to 8).

The x^16 term occurs when:
24 - 2k - 3l = 16 ==> 2k + 3l = 8.

For integers k and l such that 0 <= k, l <= 8, this occurs when:
(k, l) = (4, 0) and (1, 2).

Thus, the constant term of (x + 1/x + 1/x^2)^8 is the sum of:
C(8, k)*C(8 - k, l)

where (k, l) = (4, 0) and (1, 2).

Therefore, the required constant term is:
C(8, 4)*C(8 - 4, 0) + C(8, 1)*C(8 - 1, 2) = [8!/(4!4!)][4!/(0!4!)] + [8!/(1!7!)][7!/(2!5!)]
= (70)(1) + (8)(21)
= 238, as required.

I hope this helps!

This problem can be done and should be done lot simpler!!!
In the expansion of (x + 1/x + 1/x^2)^8, the general term contains (x)^i (1/x)^j (1/x^2)^k,
i+j+k = 8
i-j-2k = 0 (constant term requirement)
Cancel j,
k = 2i - 8
You can get two groups of integer solutions:
i = 4, j = 4, k = 0;
or
i = 5, j = 1, k = 2
By combination, the constant term = 8C4 + 8C1 * 7C2 = 238