Expression Simplification
First, simplify the expression inside the parenthesis: $ \left(\frac{x+1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x-1}{x - x^{\frac{1}{2}}}\right)^{15} $ The first term, $ \frac{x+1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} $, simplifies using the sum of cubes identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$. With $a = x^{\frac{1}{3}}$ and $b=1$, this term becomes $a+b = x^{\frac{1}{3}} + 1$. The second term, $ \frac{x-1}{x - x^{\frac{1}{2}}} $, simplifies by factoring: $ \frac{(x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}+1)}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)} = \frac{x^{\frac{1}{2}}+1}{x^{\frac{1}{2}}} = 1 + x^{-\frac{1}{2}} $. Combining these, the expression inside the parenthesis is $ (x^{\frac{1}{3}} + 1) - (1 + x^{-\frac{1}{2}}) = x^{\frac{1}{3}} - x^{-\frac{1}{2}} $.
Binomial Expansion Setup
We need the term independent of $x$ in the expansion of $ \left( x^{\frac{1}{3}} - x^{-\frac{1}{2}} \right)^{15} $. The general term $T_{r+1}$ in the binomial expansion of $ (a+b)^n $ is $ T_{r+1} = \binom{n}{r} a^{n-r} b^r $. For this problem, $n=15$, $a = x^{\frac{1}{3}}$, and $b = -x^{-\frac{1}{2}}$. $ T_{r+1} = \binom{15}{r} \left( x^{\frac{1}{3}} \right)^{15-r} \left( -x^{-\frac{1}{2}} \right)^r $ $ T_{r+1} = \binom{15}{r} (-1)^r x^{\frac{15-r}{3}} x^{-\frac{r}{2}} $ Combine the exponents of $x$: $ T_{r+1} = \binom{15}{r} (-1)^r x^{\frac{15-r}{3} - \frac{r}{2}} $
Solving for the Term Index
The term is independent of $x$ when the exponent of $x$ is zero. $ \frac{15-r}{3} - \frac{r}{2} = 0 $ Multiplying by 6 yields: $ 2(15-r) - 3r = 0 $ $ 30 - 2r - 3r = 0 $ $ 30 - 5r = 0 $ $ 5r = 30 \implies r = 6 $
Calculating the Constant Term
The term independent of $x$ corresponds to $r=6$, which is the $T_7$ term. $ T_7 = \binom{15}{6} (-1)^6 x^{\frac{15-6}{3} - \frac{6}{2}} $ $ T_7 = \binom{15}{6} \times 1 \times x^0 = \binom{15}{6} $ Calculate the binomial coefficient: $ \binom{15}{6} = \frac{15!}{6!(15-6)!} = \frac{15!}{6!9!} $ $ \binom{15}{6} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} $ Simplify the calculation: $ \binom{15}{6} = \frac{15}{5 \times 3} \times \frac{14}{2} \times 13 \times \frac{12}{6 \times 4} \times 11 \times 10 \text{ (Incorrect simplification method)} $ $ \binom{15}{6} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{720} $ Cancel terms: $ \frac{12}{6 \times 2} = 1 $ and $ \frac{15}{5 \times 3} = 1 $. $ \binom{15}{6} = \frac{1 \times 14 \times 13 \times 1 \times 11 \times 10}{4 \times 1} = \frac{14 \times 13 \times 11 \times 10}{4} $ Simplify further: $ \frac{10}{4} = \frac{5}{2} $. $ \binom{15}{6} = \frac{14 \times 13 \times 11 \times 5}{2} $ Simplify $ \frac{14}{2} = 7 $. $ \binom{15}{6} = 7 \times 13 \times 11 \times 5 = 91 \times 55 = 5005 $ The term independent of $x$ is $5005$.
Similar Questions
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- Let n be the number obtained on rolling a fair die. If the probability that the system
$x - ny + z = 6$
$x + (n - 2)y + (n + 1)z = 8$
$(n - 1)y + z = 1$
has a unique solution is $\frac{k}{6}$, then the sum of k and all possible values of n is : - If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
