Find the free coefficient of x^{-18} in the expansion of (x^3 - 1/x^4)^{15}.

Steps to Solve

1. Identifying the Binomial Expansion Terms

We want to expand the expression ((x^3 - \frac{1}{x^4})^{15}). The general term in the expansion can be represented as:

$$ T_k = \binom{n}{k} (x^3)^{n-k} \left(-\frac{1}{x^4}\right)^{k} $$

where ( n = 15 ) and ( k ) varies from 0 to 15.

2. Simplifying the General Term

Substituting ( n ) into the general term gives us:

$$ T_k = \binom{15}{k} (x^3)^{15-k} \left(-\frac{1}{x^4}\right)^{k} = \binom{15}{k} (-1)^k x^{3(15-k) - 4k} $$

This simplifies to:

$$ T_k = \binom{15}{k} (-1)^k x^{45 - 3k - 4k} = \binom{15}{k} (-1)^k x^{45 - 7k} $$

3. Finding the Required Power

We need the coefficient of ( x^{-18} ). This means we need to set the exponent equal to -18:

$$ 45 - 7k = -18 $$

4. Solving for ( k )

Now we solve the equation:

$$ 45 + 18 = 7k $$

$$ 63 = 7k $$

$$ k = 9 $$

5. Calculating the Coefficient

Now substitute ( k = 9 ) back into the general term to find the coefficient:

$$ T_9 = \binom{15}{9} (-1)^9 x^{-18} $$

So the coefficient is:

$$ \text{Coefficient} = \binom{15}{9} (-1)^9 = -\binom{15}{9} $$

6. Calculating the Value of ( \binom{15}{9} )

We can calculate it as:

$$ \binom{15}{9} = \binom{15}{6} = \frac{15!}{6! \cdot 9!} $$

Calculating this gives:

$$ \binom{15}{9} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5005 $$

Thus,

$$ \text{Coefficient} = -5005 $$