The term independent of x in the expansion of $\left( \frac{(x+1)}{(x^{2/3}+1-x^{1/3})} - \frac{(x-1)}{(x-x^{1/2})} \right)^{10}$, $x > 1$, is:
Simplifying the Expression
First, simplify the terms within the parentheses:
- Term 1: $\frac{(x+1)}{(x^{2/3}+1-x^{1/3})}$
- Let $y = x^{1/3}$. The term becomes $\frac{y^3+1}{y^2+1-y}$.
- Using the sum of cubes formula ($a^3+b^3 = (a+b)(a^2-ab+b^2)$), $y^3+1 = (y+1)(y^2-y+1)$.
- So, $\frac{(y+1)(y^2-y+1)}{y^2-y+1} = y+1 = x^{1/3}+1$.
- Term 2: $\frac{(x-1)}{(x-x^{1/2})}$
- Let $z = x^{1/2}$. The term becomes $\frac{z^2-1}{z^2-z}$.
- Factorizing: $\frac{(z-1)(z+1)}{z(z-1)} = \frac{z+1}{z} = \frac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2}$.
The expression inside the parentheses simplifies to:
$(x^{1/3}+1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$.
Binomial Expansion
The expression to expand is $(x^{1/3} - x^{-1/2})^{10}$.
Using the binomial theorem, the general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$
Here, $n=10$, $a = x^{1/3}$, $b = -x^{-1/2}$.
$T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$
$T_{r+1} = \binom{10}{r} x^{\frac{10-r}{3}} (-1)^r x^{-\frac{r}{2}}$
$T_{r+1} = \binom{10}{r} (-1)^r x^{\left(\frac{10-r}{3} - \frac{r}{2}\right)}$
Finding the Term Independent of x
For the term to be independent of $x$, the exponent of $x$ must be 0.
$ \frac{10-r}{3} - \frac{r}{2} = 0 $
Multiply by 6 to clear denominators:
$ 2(10-r) - 3r = 0 $
$ 20 - 2r - 3r = 0 $
$ 20 - 5r = 0 $
$ 5r = 20 $
$ r = 4 $
Calculating the Term
Substitute $r=4$ back into the general term formula:
$T_{4+1} = T_5 = \binom{10}{4} (-1)^4 x^{\left(\frac{10-4}{3} - \frac{4}{2}\right)}$
$T_5 = \binom{10}{4} (1) x^{\left(\frac{6}{3} - 2\right)}$
$T_5 = \binom{10}{4} x^{(2 - 2)} = \binom{10}{4} x^0$
Calculate the combination:
$ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210 $
The term independent of $x$ is 210.
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