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Find the first three terms in ascending powers of y in the expansion of (1-y)^(1/3) and the coefficient of x^(-3) in ((x+1)(2x-1/x)3).

Mathematics

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Answer

1−13y−19y21 - \frac{1}{3}y - \frac{1}{9}y^2

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3. Find the first three terms in ascending powers of yy in the expansion of (1−y)13(1-y)^{\frac{1}{3}}.

Use (1+x)n=1+nx+n(n−1)2!x2+…(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots with x=−yx=-y, n=13n=\frac{1}{3}.

  1. Calculate terms.

    11 13(−y)=−13y\frac{1}{3}(-y) = -\frac{1}{3}y 13(13−1)2!(−y)2=13(−23)2y2=−292y2=−19y2\frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}(-y)^2 = \frac{\frac{1}{3}(-\frac{2}{3})}{2}y^2 = \frac{-\frac{2}{9}}{2}y^2 = -\frac{1}{9}y^2

    The first three terms are 1−13y−19y2\boxed{1 - \frac{1}{3}y - \frac{1}{9}y^2}.

    4. Find the coefficient of x−3x^{-3} in the expansion of ((x+1)(2x−1x)3)\left((x+1)\left(2x - \frac{1}{x}\right)^3\right).

  2. Expand (2x−1x)3\left(2x - \frac{1}{x}\right)^3.

    (2x−1x)3=(2x)3+3(2x)2(−1x)+3(2x)(−1x)2+(−1x)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 =8x3−12x+6x−1−x−3= 8x^3 - 12x + 6x^{-1} - x^{-3}

  3. Multiply by (x+1)(x+1) and find x−3x^{-3} terms.

    (x+1)(8x3−12x+6x−1−x−3)(x+1)(8x^3 - 12x + 6x^{-1} - x^{-3})

    Terms contributing to x−3x^{-3}:

    • x⋅(term with x−4)x \cdot (\text{term with } x^{-4}): None in the expansion.
    • 1⋅(term with x−3)1 \cdot (\text{term with } x^{-3}): 1⋅(−x−3)=−x−31 \cdot (-x^{-3}) = -x^{-3}

  4. Identify the coefficient.

    The coefficient of x−3x^{-3} is −1-1.
    The final answer is −1\boxed{-1}.

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