Pure mathematics

In the expansion of (2x^2 + a/x)^6, the coefficients of x^6 and x^3 are equal. Find the value of the non-zero constant a and the coefficient of x^6 in the expansion of (1 - x^3)(2x^2 + a/x)^6.

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  1. Not sure just what you mean by
    In the expansion of (2x^2 + a/x)^6, the coefficients of x^6 and x^3 are equal.
    but
    (2x^2 + a/x)^6 = ... + 160a^3x^3 + 240a^2x^6 ...

    On the other hand,
    (1 - x^3)(2x^2 + a/x)^6 = ... - 160a^3x^6 + 240a^2x^6

    Maybe that will mean something to you.

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    oobleck
  2. In the expansion, oobleck gave you the terms containing x^6 and x^3, namely 160a^3x^3 and 240a^2x^6
    If their coefficients are equal,
    160a^3 = 240a^2
    160a^3 - 240a^2= 0
    a^2(160a - 240) = 0
    a = 0, or a = 240/160 = 3/2
    if a = 0, you would simply have 64x^12, since all terms containing "a" would drop out.

    In the expansion of (1 - x^3)(2x^2 + a/x)^6, the terms that would give you x^6 are:
    1(240a^2x^6) and (-x^3)(160a^3x^3)

    then
    1(240a^2x^6) + (-x^3)(160a^3x^3)
    = 240(9/4)x^6 - 160(27/8)x^6
    = 540x^6 - 540x^6
    = 0

    so the coefficient of that term is zero

    notice in
    www.wolframalpha.com/input/?i=expand+%281-x%5E3%29%282x%5E2+%2B+%283%2F2%29%2Fx%29%5E6

    that we don't have an x^6 term

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