Expanding the trinomial, (x+y+z)^n, gives terms X^rY^sZ^t where r+s+t=n and with a coeficient of n! over r!s!t!.

a. use this pattern to expand (x+y+z)^3.
b. find the sum of the coeficients in the expression of (x+y+z)^n.

(LET x=y=z=1.)

To expand the trinomial (x+y+z)^n, we can use the pattern you mentioned. The terms of the expansion will have the form X^rY^sZ^t, where r, s, and t are non-negative integers such that r + s + t = n. The coefficient of each term will be n! / (r!s!t!).

a. To expand (x+y+z)^3 using this pattern:
We can substitute n = 3 into the formula. So the terms in the expansion will have the form X^rY^sZ^t, where r + s + t = 3. We need to find all possible combinations of r, s, and t that satisfy this equation.

The combinations of r, s, and t are:
(3, 0, 0)
(2, 1, 0)
(2, 0, 1)
(1, 2, 0)
(1, 1, 1)
(1, 0, 2)
(0, 3, 0)
(0, 2, 1)
(0, 1, 2)
(0, 0, 3)

For each combination, we can calculate the corresponding term of the expansion:
(3, 0, 0): x^3
(2, 1, 0): x^2y
(2, 0, 1): x^2z
(1, 2, 0): xy^2
(1, 1, 1): xyz
(1, 0, 2): xz^2
(0, 3, 0): y^3
(0, 2, 1): y^2z
(0, 1, 2): yz^2
(0, 0, 3): z^3

Now we can calculate their coefficients:
For each combination (r, s, t), we can calculate the coefficient using n! / (r!s!t!):
(3, 0, 0): 3! / (3!0!0!) = 1
(2, 1, 0): 3! / (2!1!0!) = 3
(2, 0, 1): 3! / (2!0!1!) = 3
(1, 2, 0): 3! / (1!2!0!) = 3
(1, 1, 1): 3! / (1!1!1!) = 6
(1, 0, 2): 3! / (1!0!2!) = 3
(0, 3, 0): 3! / (0!3!0!) = 1
(0, 2, 1): 3! / (0!2!1!) = 3
(0, 1, 2): 3! / (0!1!2!) = 3
(0, 0, 3): 3! / (0!0!3!) = 1

Putting it all together, the expansion of (x+y+z)^3 is:
x^3 + 3x^2y + 3x^2z + 3xy^2 + 6xyz + 3xz^2 + y^3 + 3y^2z + 3yz^2 + z^3

b. To find the sum of the coefficients in the expression of (x+y+z)^n, we can evaluate the expression by substituting x=y=z=1:

If we substitute x=y=z=1 in each term of the above expansion, we get:
1^3 + 3(1^2) + 3(1^2) + 3(1^2) + 6(1^3) + 3(1^2) + 1^3 + 3(1^2) + 3(1^2) + 1^3

Simplifying, we have:
1 + 3 + 3 + 3 + 6 + 3 + 1 + 3 + 3 + 1

Adding all the terms, we get the sum of the coefficients:
1 + 3 + 3 + 3 + 6 + 3 + 1 + 3 + 3 + 1 = 27

Therefore, the sum of the coefficients in the expression of (x+y+z)^n, when x=y=z=1, is 27.