Which term contains `x^8` in the expansion of `(x^2 - 1/x)^10`? Also , find its coefficient.

Answer

To find the term that contains \( x^8 \) in the expansion of \( (x^2 - \frac{1}{x})^{10} \), we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, let \( a = x^2 \) and \( b = -\frac{1}{x} \), and \( n = 10 \).

### Step 1: Write the general term
The general term \( T_{r+1} \) in the expansion is given by:

\[ T_{r+1} = \binom{10}{r} (x^2)^{10-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, simplifying \( T_{r+1} \):

\[ T_{r+1} = \binom{10}{r} (x^2)^{10-r} \cdot (-1)^r \cdot \frac{1}{x^r} \]
\[ = \binom{10}{r} (-1)^r x^{2(10-r)} \cdot x^{-r} \]
\[ = \binom{10}{r} (-1)^r x^{20 - 2r - r} \]
\[ = \binom{10}{r} (-1)^r x^{20 - 3r} \]

### Step 3: Set the exponent of \( x \) equal to 8
We need the exponent of \( x \) to be 8:

\[ 20 - 3r = 8 \]

### Step 4: Solve for \( r \)
Rearranging the equation:

\[ 20 - 8 = 3r \]
\[ 12 = 3r \]
\[ r = 4 \]

### Step 5: Find the term corresponding to \( r = 4 \)
Now we substitute \( r = 4 \) back into the general term:

\[ T_{5} = \binom{10}{4} (-1)^4 x^{20 - 3 \cdot 4} \]
\[ = \binom{10}{4} x^{8} \]

### Step 6: Calculate the coefficient
Now we calculate the coefficient:

\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

### Conclusion
Thus, the term containing \( x^8 \) is \( 210 x^8 \) and the coefficient is \( 210 \).

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