To expand the expressions, we will use the binomial theorem, which states that for any non-negative integer n:

To expand the expressions, we will use the binomial theorem, which states that for any non-negative integer n: (a+b)^n = _k=0^n nk a^n-k b^k The coefficients nk for n=5 are found in Row 5 of Pascal's triangle: 1, 5, 10, 10, 5, 1. Part 1: Expand (x+6y)^5 Here, a=x and b=6y. Step 1: Apply the binomial theorem with n=5. (x+6y)^5 = 50x^5(6y)^0 + 51x^4(6y)^1 + 52x^3(6y)^2 + 53x^2(6y)^3 + 54x^1(6y)^4 + 55x^0(6y)^5 Step 2: Substitute the binomial coefficients and simplify each term. = 1 · x^5 · 1 + 5 · x^4 · (6y) + 10 · x^3 · (36y^2) + 10 · x^2 · (216y^3) + 5 · x · (1296y^4) + 1 · 1 · (7776y^5) Step 3: Perform the multiplications. = x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5 The expansion of (x+6y)^5 is: x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5 Part 2: Expand (3-6y)^5 Here, a=3 and b=-6y. Step 1: Apply the binomial theorem with n=5. (3-6y)^5 = 50(3)^5(-6y)^0 + 51(3)^4(-6y)^1 + 52(3)^3(-6y)^2 + 53(3)^2(-6y)^3 + 54(3)^1(-6y)^4 + 55(3)^0(-6y)^5 Step 2: Substitute the binomial coefficients and simplify each term. = 1 · (243) · 1 + 5 · (81) · (-6y) + 10 · (27) · (36y^2) + 10 · (9) · (-216y^3) + 5 · (3) · (1296y^4) + 1 · 1 · (-7776y^5) Step 3: Perform the multiplications. = 243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5 The expansion of (3-6y)^5 is: 243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5 That's 2 down. 3 left today — send the next one.