Problem Analysis

We need to find the value of $a$ such that the coefficient of $x^3$ in the expansion of $(3 + ax)^6$ equals $160$.

The general term in the binomial expansion of $(3 + ax)^6$ is given by:

$T_r = \binom{6}{r} (3)^{6-r} (ax)^r$

The term containing $x^3$ corresponds to $r = 3$. Let's compute it step by step.

Step 1: Write the term for $r = 3$

Substitute $r = 3$ into the general term formula:

$T_3 = \binom{6}{3} (3)^{6-3} (ax)^3$

Step 2: Simplify the coefficients

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = 20$ $(3)^{6-3} = 3^3 = 27$ $(ax)^3 = a^3 x^3$

Thus, the term becomes: $T_3 = 20 \cdot 27 \cdot a^3 x^3$

$T_3 = 540 a^3 x^3$

Step 3: Compare coefficients

The coefficient of $x^3$ is $540a^3$. We are given that this coefficient is equal to $160$. So, set up the equation: $540a^3 = 160$

Step 4: Solve for $a$

$a^3 = \frac{160}{540} = \frac{8}{27}$ $a = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$

Final Answer:

$a = \frac{2}{3}$

Let me know if you'd like further clarifications or more details! 😊

Related Questions:

1. What is the general formula for the $r$-th term in a binomial expansion?
2. How does the value of $a$ affect the coefficient of $x^3$?
3. What are some other properties of binomial coefficients?
4. Can you derive the term for $x^4$ in the same expansion?
5. How would the solution change if the exponent were different (e.g., $(3 + ax)^5$)?

Tip:

When solving binomial expansion problems, focus on identifying the term with the specific power of $x$ you're interested in. This simplifies calculations significantly!