3. Find the first three terms in ascending powers of y in the expansion of (1-y)^(1)/(3).
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions.
3. Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{3}}$.
Use the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$
Here $x=-y$ and $n=\frac{1}{3}$.
Step 1: First term.
$$ 1 $$
Step 2: Second term.
$$ n(-y) = \frac{1}{3}(-y) = -\frac{1}{3}y $$
Step 3: Third term.
$$ \frac{n(n-1)}{2!}(-y)^2 = \frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2} y^2 = \frac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2} y^2 = \frac{-\frac{2}{9}}{2} y^2 = -\frac{1}{9}y^2 $$
The first three terms are $\boxed{1 - \frac{1}{3}y - \frac{1}{9}y^2}$.
4. Find the coefficient of $x^{-3}$ in the expansion of $\left((x+1)\left(2x - \frac{1}{x}\right)^3\right)$.
Step 1: Expand $\left(2x - \frac{1}{x}\right)^3$.
$$ \left(2x - \frac{1}{x}\right)^3 = (2x)^3 + 3(2x)^2\left(-\frac{1}{x}\right) + 3(2x)\left(-\frac{1}{x}\right)^2 + \left(-\frac{1}{x}\right)^3 $$
$$ = 8x^3 + 3(4x^2)\left(-\frac{1}{x}\right) + 3(2x)\left(\frac{1}{x^2}\right) - \frac{1}{x^3} $$
$$ = 8x^3 - 12x + \frac{6x}{x^2} - \frac{1}{x^3} $$
$$ = 8x^3 - 12x + 6x^{-1} - x^{-3} $$
Step 2: Multiply by $(x+1)$.
$$ (x+1)(8x^3 - 12x + 6x^{-1} - x^{-3}) $$
$$ = x(8x^3 - 12x + 6x^{-1} - x^{-3}) + 1(8x^3 - 12x + 6x^{-1} - x^{-3}) $$
$$ = (8x^4 - 12x^2 + 6x^0 - x^{-2}) + (8x^3 - 12x + 6x^{-1} - x^{-3}) $$
Step 3: Identify the coefficient of $x^{-3}$.
The only term with $x^{-3}$ is $-x^{-3}$.
The coefficient of $x^{-3}$ is $-1$.
The final answer is $\boxed{-1}$.
