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Find the binomial expansion of x2(x-2y)3. 2. Find the term independent of x in the expansion of (x - (3)/(x))4. 3. Find the first three terms in ascending powers of y in the expansion of (1-y)^(1)/(3)

Mathematics

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1. Find the binomial expansion of x2(x−2y)3x^2(x-2y)^3.

  1. Expand (x−2y)3(x-2y)^3 using the binomial theorem (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.

    Here a=xa=x, b=−2yb=-2y, n=3n=3.

    (x−2y)3=(30)x3(−2y)0+(31)x2(−2y)1+(32)x1(−2y)2+(33)x0(−2y)3(x-2y)^3 = \binom{3}{0}x^3(-2y)^0 + \binom{3}{1}x^2(-2y)^1 + \binom{3}{2}x^1(-2y)^2 + \binom{3}{3}x^0(-2y)^3 =1⋅x3⋅1+3⋅x2⋅(−2y)+3⋅x⋅(4y2)+1⋅1⋅(−8y3)= 1 \cdot x^3 \cdot 1 + 3 \cdot x^2 \cdot (-2y) + 3 \cdot x \cdot (4y^2) + 1 \cdot 1 \cdot (-8y^3) =x3−6x2y+12xy2−8y3= x^3 - 6x^2y + 12xy^2 - 8y^3
  2. Multiply the result by x2x^2.

    x2(x3−6x2y+12xy2−8y3)=x5−6x4y+12x3y2−8x2y3x^2(x^3 - 6x^2y + 12xy^2 - 8y^3) = x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3

    The final answer is x5−6x4y+12x3y2−8x2y3\boxed{x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3}.

    2. Find the term independent of xx in the expansion of (x−3x)4\left(x - \frac{3}{x}\right)^4.

  3. Write the general term Tk+1T_{k+1} for the expansion of (a+b)n(a+b)^n.

    Here a=xa=x, b=−3xb=-\frac{3}{x}, n=4n=4.

    Tk+1=(4k)(x)4−k(−3x)kT_{k+1} = \binom{4}{k} (x)^{4-k} \left(-\frac{3}{x}\right)^k =(4k)x4−k(−3)kx−k= \binom{4}{k} x^{4-k} (-3)^k x^{-k} =(4k)(−3)kx4−2k= \binom{4}{k} (-3)^k x^{4-2k}
  4. For the term independent of xx, set the power of xx to 0.

    4−2k=0  ⟹  2k=4  ⟹  k=24-2k = 0 \implies 2k = 4 \implies k=2
  5. Substitute k=2k=2 into the general term.

    T2+1=(42)(−3)2x4−2(2)T_{2+1} = \binom{4}{2} (-3)^2 x^{4-2(2)} =6⋅9⋅x0= 6 \cdot 9 \cdot x^0 =54= 54

    The final answer is 54\boxed{54}.

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

    Use the binomial expansion for fractional powers: (1+x)n=1+nx+n(n−1)2!x2+…(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots
    Here x=−yx=-y and n=13n=\frac{1}{3}.

  6. First term.

    11
  7. Second term.

    n(−y)=13(−y)=−13yn(-y) = \frac{1}{3}(-y) = -\frac{1}{3}y
  8. Third term.

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

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

    Let a=2xa=2x and b=−1xb=-\frac{1}{x}.

    (2x−1x)3=(30)(2x)3+(31)(2x)2(−1x)+(32)(2x)(−1x)2+(33)(−1x)3(2x - \frac{1}{x})^3 = \binom{3}{0}(2x)^3 + \binom{3}{1}(2x)^2\left(-\frac{1}{x}\right) + \binom{3}{2}(2x)\left(-\frac{1}{x}\right)^2 + \binom{3}{3}\left(-\frac{1}{x}\right)^3 =1⋅8x3+3⋅4x2⋅(−1x)+3⋅2x⋅(1x2)+1⋅(−1x3)= 1 \cdot 8x^3 + 3 \cdot 4x^2 \cdot \left(-\frac{1}{x}\right) + 3 \cdot 2x \cdot \left(\frac{1}{x^2}\right) + 1 \cdot \left(-\frac{1}{x^3}\right) =8x3−12x+6x−1x3= 8x^3 - 12x + \frac{6}{x} - \frac{1}{x^3}
  10. Multiply the expansion by (x+1)(x+1).

    (x+1)(8x3−12x+6x−1x3)(x+1)\left(8x^3 - 12x + \frac{6}{x} - \frac{1}{x^3}\right) =x(8x3−12x+6x−1x3)+1(8x3−12x+6x−1x3)= x\left(8x^3 - 12x + \frac{6}{x} - \frac{1}{x^3}\right) + 1\left(8x^3 - 12x + \frac{6}{x} - \frac{1}{x^3}\right) =(8x4−12x2+6−x−2)+(8x3−12x+6x−1−x−3)= (8x^4 - 12x^2 + 6 - x^{-2}) + (8x^3 - 12x + 6x^{-1} - x^{-3})
  11. Identify the term with x−3x^{-3}.

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

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