Write the coefficient of x5in the expansion of: (a) (3 + x)6 (b) (2 − x)7 (c)

(a) The coefficient of x^5 in the expansion of (3 + x)^6 can be found using the binomial theorem, which states that (a + b)^n = Σ (n choose k) * a^(n-k) * b^k, where the sum is from k = 0 to n.

In this case, a = 3, b = x, and n = 6. We want the term where k = 5, so we plug these values into the formula to get:

(6 choose 5) * 3^(6-5) * x^5 = 6 * 3 * x^5 = 18x^5

So, the coefficient of x^5 is 18.

(b) The coefficient of x^5 in the expansion of (2 - x)^7 can be found in a similar way. However, because of the negative sign, we need to take into account that the sign of the term will alternate.

In this case, a = 2, b = -x, and n = 7. We want the term where k = 5, so we plug these values into the formula to get:

(7 choose 5) * 2^(7-5) * (-x)^5 = 21 * 4 * -x^5 = -84x^5

So, the coefficient of x^5 is -84.

(c) You didn't provide a third expression, so I can't find the coefficient of x^5 for it.