To expand the expressions, we will use the binomial theorem, which states that for any non-negative integer $n$: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ The coefficients $\binom{n}{k}$ 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 $n=5$. $$(x+6y)^5 = \binom{5}{0}x^5(6y)^0 + \binom{5}{1}x^4(6y)^1 + \binom{5}{2}x^3(6y)^2 + \binom{5}{3}x^2(6y)^3 + \binom{5}{4}x^1(6y)^4 + \binom{5}{5}x^0(6y)^5$$ Step 2: Substitute the binomial coefficients and simplify each term. $$ = 1 \cdot x^5 \cdot 1 + 5 \cdot x^4 \cdot (6y) + 10 \cdot x^3 \cdot (36y^2) + 10 \cdot x^2 \cdot (216y^3) + 5 \cdot x \cdot (1296y^4) + 1 \cdot 1 \cdot (7776y^5)$$ 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: $$\boxed{x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5}$$ Part 2: Expand $(3-6y)^5$ Here, $a=3$ and $b=-6y$. Step 1: Apply the binomial theorem with $n=5$. $$(3-6y)^5 = \binom{5}{0}(3)^5(-6y)^0 + \binom{5}{1}(3)^4(-6y)^1 + \binom{5}{2}(3)^3(-6y)^2 + \binom{5}{3}(3)^2(-6y)^3 + \binom{5}{4}(3)^1(-6y)^4 + \binom{5}{5}(3)^0(-6y)^5$$ Step 2: Substitute the binomial coefficients and simplify each term. $$ = 1 \cdot (243) \cdot 1 + 5 \cdot (81) \cdot (-6y) + 10 \cdot (27) \cdot (36y^2) + 10 \cdot (9) \cdot (-216y^3) + 5 \cdot (3) \cdot (1296y^4) + 1 \cdot 1 \cdot (-7776y^5)$$ Step 3: Perform the multiplications. $$ = 243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5$$ The expansion of $(3-6y)^5$ is: $$\boxed{243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5}$$ That's 2 down. 3 left today — send the next one.