The image you uploaded shows a binomial expansion problem where you are asked to find the two possible values of the constant $a$ given that the coefficient of $x^3$ in the expansion of $(2 + a x)^6$ is 60.

We can solve this problem using the binomial theorem. Let's break it down:

Step 1: General form of the binomial expansion

The binomial expansion of $(2 + ax)^6$ can be written as:

$(2 + ax)^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (a x)^k$

The general term in this expansion is:

$\binom{6}{k} (2)^{6-k} a^k x^k$

Step 2: Finding the term for $x^3$

We need the term where $k = 3$ because we are looking for the coefficient of $x^3$. For $k = 3$, the term becomes:

$\binom{6}{3} (2)^{6-3} a^3 x^3 = \binom{6}{3} (2)^3 a^3 x^3$

Step 3: Simplifying the expression

First, calculate $\binom{6}{3}$ and $2^3$:

$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

$2^3 = 8$

Thus, the term for $x^3$ is:

$20 \times 8 \times a^3 x^3 = 160 a^3 x^3$

Step 4: Setting the coefficient equal to 60

You are told that the coefficient of $x^3$ is 60. Therefore, we set:

$160 a^3 = 60$

Step 5: Solving for $a$

Solve the equation $160 a^3 = 60$:

$a^3 = \frac{60}{160} = \frac{3}{8}$

Now, take the cube root of both sides:

[ a = \sqrt[3]{\frac{3}{8}} = \frac{\sqrt[3]{3}}{\sqrt[3]{8}} = \frac{\sqrt[3]{3}}{2} ]

So, the two possible values for $a$ are:

[ a = \frac{\sqrt[3]{3}}{2} \quad \text{or} \quad a = -\frac{\sqrt[3]{3}}{2} ]

Conclusion:

The two possible values for $a$ are $\frac{\sqrt[3]{3}}{2}$ and $-\frac{\sqrt[3]{3}}{2}$.

Would you like further details on the steps, or do you have any questions about this solution?

Here are some related questions:

1. How do you generalize binomial expansions for other powers?
2. Can the binomial theorem be applied to negative exponents?
3. How do you find the coefficient of a general term in a binomial expansion?
4. What is the significance of choosing the right value of $k$ in binomial expansions?
5. How would the solution change if the exponent was different, say 7 instead of 6?

Tip: In binomial expansions, the power of $x$ in each term corresponds to the index $k$. This can be used to directly identify which term to focus on for a given power of $x$.