To find the term independent of $$x$$ in the expansion, we first simplify the two fractions inside the parentheses.
Step 1: Simplify the First Fraction
Let the first term be $$A$$:
$$A = \frac{x+1}{x^{2/3} - x^{1/3} + 1}$$
Using the sum of cubes factorization $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$, we can write $$x + 1$$ as:
$$x + 1 = (x^{1/3})^3 + 1^3 = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$$
Substituting this back into $$A$$:
$$A = \frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$
Step 2: Simplify the Second Fraction
Let the second term be $$B$$:
$$B = \frac{x-1}{x - x^{1/2}}$$
Using the difference of squares $$a^2 - b^2 = (a - b)(a + b)$$ and factoring the denominator:
- Numerator: $$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1)$$
- Denominator: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$
Substituting these back into $$B$$:
$$B = \frac{(x^{1/2} - 1)(x^{1/2} + 1)}{x^{1/2}(x^{1/2} - 1)} = \frac{x^{1/2} + 1}{x^{1/2}} = 1 + x^{-1/2}$$
Step 3: Set up the Binomial Expansion
Now, we subtract the simplified terms:
$$A - B = (x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$$
We need to find the term independent of $x$ in the expansion of:
$$(x^{1/3} - x^{-1/2})^{10}$$
The general term $$T_{r+1}$$ is given by:
$$T_{r+1} = \binom{10}{r} \left(x^{1/3}\right)^{10-r} \left(-x^{-1/2}\right)^r$$ $$T_{r+1} = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$$
Step 4: Find $$r$$ and Compute the Term
For the term to be independent of $$x$$, the exponent of $$x$$ must be $$0$$:
$$\frac{10-r}{3} - \frac{r}{2} = 0$$
Multiply the entire equation by $$6$$ to clear the denominators:
$$2(10-r) - 3r = 0$$ $$20 - 2r - 3r = 0$$ $$5r = 20 \implies r = 4$$
Now substitute $$r = 4$$ back into the general term formula:
$$T_{5} = \binom{10}{4} (-1)^4 = \binom{10}{4}$$ $$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
Final Answer
The term independent of $$x$$ is $$210$$.
91- 6303239042
