The tenth term of the expansion (x² - 1/x)^{12}

Steps to Solve

1. Identify the Binomial Coefficient
   To find the tenth term of the expansion, we need the binomial coefficient. The general term in the binomial expansion is given by:
   $$ T_{k+1} = \binom{n}{k} a^{n-k} b^k $$
   For our expression, ( n = 12 ) and for the tenth term, ( k = 9 ).

2. Determine 'a' and 'b'
   In the expression ( (x^2 - \frac{1}{x})^{12} ):

• ( a = x^2 )
• ( b = -\frac{1}{x} )

3. Calculate the Tenth Term
   Using the formula for the general term, we substitute ( n ), ( k ), ( a ), and ( b ):
   $$ T_{10} = \binom{12}{9} (x^2)^{12-9} \left(-\frac{1}{x}\right)^9 $$
   This simplifies to:
   $$ T_{10} = \binom{12}{9} (x^2)^3 \left(-\frac{1}{x}\right)^9 $$

4. Calculate Each Component

• The binomial coefficient:
  $$ \binom{12}{9} = \binom{12}{3} = \frac{12!}{3!(12-3)!} = 220 $$
• Next, simplify ( (x^2)^3 = x^6 )
• And ( \left(-\frac{1}{x}\right)^9 = -\frac{1}{x^9} ) adds up to ( -\frac{1}{x^9} )

5. Combine the Components
   Now, substitute the values back into the term:
   $$ T_{10} = 220 \cdot x^6 \cdot \left(-\frac{1}{x^9}\right) $$
   This simplifies to:
   $$ T_{10} = -220 \cdot \frac{x^6}{x^9} = -220 \cdot x^{-3} $$