Here's a step-by-step solution to the problem. I will assume that 'z' in the first part of the question is a typo and should be 'x', consistent with the second part of the question. Part 1: Find the value of the constant $a$. We are given that the coefficient of $x^3$ in the expansion of $(3+ax)^6$ is 160. Step 1: Write the general term of the binomial expansion. The general term in the expansion of $(A+B)^n$ is given by $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. For $(3+ax)^6$, we have $A=3$, $B=ax$, and $n=6$. So, the general term is: $$ T_{r+1} = \binom{6}{r} (3)^{6-r} (ax)^r $$ $$ T_{r+1} = \binom{6}{r} 3^{6-r} a^r x^r $$ Step 2: Find the term containing $x^3$. For the term with $x^3$, we set $r=3$: $$ T_{3+1} = T_4 = \binom{6}{3} 3^{6-3} a^3 x^3 $$ $$ T_4 = \binom{6}{3} 3^3 a^3 x^3 $$ Step 3: Calculate the binomial coefficient and simplify. $$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 $$ $$ 3^3 = 27 $$ Substitute these values back into the expression for $T_4$: $$ T_4 = 20 \times 27 \times a^3 x^3 $$ $$ T_4 = 540 a^3 x^3 $$ The coefficient of $x^3$ is $540a^3$. Step 4: Solve for $a$. We are given that the coefficient of $x^3$ is 160: $$ 540 a^3 = 160 $$ $$ a^3 = \frac{160}{540} $$ $$ a^3 = \frac{16}{54} $$ $$ a^3 = \frac{8}{27} $$ Take the cube root of both sides: $$ a = \sqrt[3]{\frac{8}{27}} $$ $$ a = \frac{2}{3} $$ The value of the constant $a$ is $\boxed{\frac{2}{3}}$. Part 2: Find the coefficient of $x^3$ in the expansion of $(3+ax)^6(1-2x)$. Now we use the value $a=\frac{2}{3}$. The expression becomes $(3+\frac{2}{3}x)^6(1-2x)$. Step 1: Expand $(3+\frac{2}{3}x)^6$ up to the $x^3$ term. We need the coefficients of $x^0$, $x^1$, $x^2$, and $x^3$ from $(3+\frac{2}{3}x)^6$. Using the general term $T_{r+1} = \binom{6}{r} 3^{6-r} (\frac{2}{3}x)^r = \binom{6}{r} 3^{6-r} (\frac{2}{3})^r x^r$: Coefficient of $x^0$ (constant term, $r=0$): $$ \binom{6}{0} 3^6 (\frac{2}{3})^0 = 1 \times 729 \times 1 = 729 $$ Coefficient of $x^1$ ($r=1$): $$ \binom{6}{1} 3^5 (\frac{2}{3})^1 = 6 \times 243 \times \frac{2}{3} = 6 \times 81 \times 2 = 972 $$ Coefficient of $x^2$ ($r=2$): $$ \binom{6}{2} 3^4 (\frac{2}{3})^2 = 15 \times 81 \times \frac{4}{9} = 15 \times 9 \times 4 = 540 $$ Coefficient of $x^3$ ($r=3$): $$ \binom{6}{3} 3^3 (\frac{2}{3})^3 = 20 \times 27 \times \frac{8}{27} = 20 \times 8 = 160 $$ So, $(3+\frac{2}{3}x)^6 = 729 + 972x + 540x^2 + 160x^3 + \dots$ Step 2: Multiply the expansion by $(1-2x)$ and identify $x^3$ terms. We need the coefficient of $x^3$ in $(729 + 972x + 540x^2 + 160x^3 + \dots)(1-2x)$. To get an $x^3$ term, we can multiply: The $x^3$ term from the first bracket by the constant term from the second bracket: $(160x^3) \times (1) = 160x^3$ The $x^2$ term from the first bracket by the $x$ term from the second bracket: $(540x^2) \times (-2x) = -1080x^3$ Step 3: Sum the coefficients of the $x^3$ terms. The total coefficient of $x^3$ is the sum of these contributions: $$ 160 + (-1080) = 160 - 1080 = -920 $$ The coefficient of $x^3$ in the