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The image shows a mathematical problem where we are asked to find the term free from \( x \) in the expansion of the binomial expression:

\[
\left( x^2 - \frac{6}{x} \right)^9
\]

To solve this, we will expand the binomial expression using the binomial theorem, which is given by:

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

In this case, \( a = x^2 \), \( b = -\frac{6}{x} \), and \( n = 9 \).

The general term in the expansion of \( (x^2 - \frac{6}{x})^9 \) is:

\[
T_k = \binom{9}{k} (x^2)^{9-k} \left( -\frac{6}{x} \right)^k
\]

Simplifying this expression:

\[
T_k = \binom{9}{k} x^{2(9-k)} (-1)^k \frac{6^k}{x^k}
\]

This further simplifies to:

\[
T_k = \binom{9}{k} (-1)^k 6^k x^{18 - 3k}
\]

Now, we are asked to find the term free from \( x \). For a term to be independent of \( x \), the exponent of \( x \) must be zero. Therefore, we set the exponent of \( x \) in the general term to zero:

\[
18 - 3k = 0
\]

Solving for \( k \):

\[
3k = 18 \quad \Rightarrow \quad k = 6
\]

So, the term free from \( x \) occurs when \( k = 6 \). Substituting \( k = 6 \) into the general term formula:

\[
T_6 = \binom{9}{6} (-1)^6 6^6
\]

Since \( (-1)^6 = 1 \), we have:

\[
T_6 = \binom{9}{6} 6^6
\]

Now, calculate \( \binom{9}{6} \), which is equal to \( \binom{9}{3} \) because of symmetry:

\[
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]

Thus, the term free from \( x \) is:

\[
T_6 = 84 \times 6^6
\]

Finally, calculate \( 6^6 \):

\[
6^6 = 46656
\]

Therefore, the term free from \( x \) is:

\[
T_6 = 84 \times 46656 = 3919104
\]

So, the constant term is \( 3,919,104 \).

Would you like further details on this solution?

Here are 5 related questions:
1. How does the binomial theorem apply to polynomial expansions?
2. What is the role of the general term in identifying specific terms in an expansion?
3. Why is the exponent of \( x \) set to zero to find the constant term?
4. How can you compute binomial coefficients using Pascal’s triangle?
5. Can the same process be applied to other powers or terms in binomial expansions?

**Tip:** Binomial expansions can often simplify complex algebraic expressions and are particularly useful when looking for specific terms or behaviors of a function.

In the expansion of (x^2 - 6/x)^9, find the term free from x.

This problem covers how to find the term free from x in the expansion of (x^2 - 6/x)^9 using the binomial theorem. Detailed steps are provided to simplify the binomial expansion, including exponent rules and solving for the constant term. Suitable for students in Grades 10-12.

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