Give that coefficient of x^3 in the expansion of [a+x^3 \2 ]^4 is 1.Find the value of a

You're on a roll — Step 1: Write the general term for the binomial expansion of $\left(a + \frac{x}{2}\right)^4$. The general term $T_{r+1}$ is given by: $$ T_{r+1} = \binom{n}{r} A^{n-r} B^r $$ In this case, $A=a$, $B=\frac{x}{2}$, and $n=4$. $$ T_{r+1} = \binom{4}{r} a^{4-r} \left(\frac{x}{2}\right)^r $$ $$ T_{r+1} = \binom{4}{r} a^{4-r} \frac{x^r}{2^r} $$ Step 2: Find the term containing $x^3$. For the term containing $x^3$, we need $r=3$. Substitute $r=3$ into the general term: $$ T_{3+1} = \binom{4}{3} a^{4-3} \frac{x^3}{2^3} $$ $$ T_4 = \binom{4}{3} a^1 \frac{x^3}{8} $$ Step 3: Calculate the binomial coefficient $\binom{4}{3}$. $$ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4 $$ Step 4: Substitute the value of $\binom{4}{3}$ back into the term. $$ T_4 = 4 \cdot a \cdot \frac{x^3}{8} $$ $$ T_4 = \frac{4a}{8} x^3 $$ $$ T_4 = \frac{a}{2} x^3 $$ Step 5: Identify the coefficient of $x^3$ and set it equal to 1. The coefficient of $x^3$ is $\frac{a}{2}$. Given that the coefficient of $x^3$ is 1: $$ \frac{a}{2} = 1 $$ Step 6: Solve for $a$. $$ a = 1 \times 2 $$ $$ a = 2 $$ The value of $a$ is $\boxed{2}$. What's next?