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Math Help - Expanding binomial to a fraction power

  1. #1
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    Expanding binomial to a fraction power

    I have the following:

    <br />
(x^4 - 1)^\frac{1}{3}

    I know you can make that into a root:

    <br />
\sqrt[3]{x^4 -1}

    but is there a way to expend that out, like you would expand:

    <br />
(x^4 - 1)^2 = x^8 - 2x^4 + 1

    thanks
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  2. #2
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    Sure. Just do it...

    (x^4 - 1)^(1/3) = (x^4)^{1/3} + (1/3)

     <br />
(x^{4}-1)^{\frac{1}{3}} = \frac{1}{0!}(x^4)^{\frac{1}{3}}(-1)^{0} + \frac{\frac{1}{3}}{1!}(x^4)^{\frac{1}{3}-1}(-1)^{1} + \frac{\frac{1}{3} \cdot \frac{-2}{3}}{2!}(x^4)^{\frac{1}{3}-2}(-1)^{2} + ...<br />

    It is a wonderfully useful result.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Sure. Just do it...

    (x^4 - 1)^(1/3) = (x^4)^{1/3} + (1/3)

     <br />
(x^{4}-1)^{\frac{1}{3}} = \frac{1}{0!}(x^4)^{\frac{1}{3}}(-1)^{0} + \frac{\frac{1}{3}}{1!}(x^4)^{\frac{1}{3}-1}(-1)^{1} + \frac{\frac{1}{3} \cdot \frac{-2}{3}}{2!}(x^4)^{\frac{1}{3}-2}(-1)^{2} + ...<br />

    It is a wonderfully useful result.
    Wait, how'd you go from:

    <br />
(x^4 - 1)^\frac{1}{3} = (x^4)^\frac{1}{3} + (\frac{1}{3})

    to all that stuff at the bottom?
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  4. #4
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    It is only a minor generalization from the finite Binomial Expansion

    (a + b)^{n} = a^{n} + ... See how it starts out with the first tern to the external exponent?

    (a + b)^{n} = a^{n} + n*a^{n-1}b + ... Then we start reducing the exponent on the first term and increasing the exponent on the second term. The rest is just controlling the combinations.
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