Step 1: Determine the formula for the coefficient of a term in a binomial expansion. The $(r+1)$-th term in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. The coefficient of the $(r+1)$-th term is $\binom{n}{r}$. Step 2: Find the coefficients of the sixth and sixteenth terms. For the sixth term, $r+1=6$, so $r=5$. The coefficient is $\binom{n}{5}$. For the sixteenth term, $r+1=16$, so $r=15$. The coefficient is $\binom{n}{15}$. Step 3: Set the coefficients equal and solve for $n$. Given that the coefficients are equal: $$ \binom{n}{5} = \binom{n}{15} $$ According to the property of binomial coefficients, if $\binom{n}{r} = \binom{n}{k}$, then either $r=k$ or $r+k=n$. Since $5 \neq 15$, we must have $5+15=n$. $$ n = 5+15 $$ $$ n = 20 $$ The value of $n$ is: $$ \boxed{20} $$ Step 1: Group the terms to apply the binomial theorem. Let $X = (a+b)$. The expression becomes $(X-c)^3$. Step 2: Expand $(X-c)^3$ using the binomial theorem. The binomial theorem states $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$. Here, $A=X$ and $B=-c$. $$ (X-c)^3 = X^3 + 3X^2(-c) + 3X(-c)^2 + (-c)^3 $$ $$ = X^3 - 3cX^2 + 3c^2X - c^3 $$ Step 3: Substitute $X=(a+b)$ back into the expanded expression. $$ (a+b-c)^3 = (