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Math Help - [SOLVED] Check this algebric operation

  1. #1
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    Question [SOLVED] Check this algebric operation

    (a^\frac {1}{3}-b^\frac {1}{3}) (a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3})

    (a^{\frac {1}{3}+\frac {2}{3}})+(a^{\frac {1}{3}+\frac {1}{3}}b^{\frac {1}{3}})-(b^{\frac {1}{3}}a^{\frac {2}{3}})-(b^{\frac {1}{3}+\frac {1}{3}}a^{\frac {1}{3}})-(b^{\frac {1}{3}+\frac {2}{3}})<br />

    <br />
(a^{\frac {3}{3}})+(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {1}{3}}b^\frac {2}{3})-(b^{\frac {3}{3}})<br />



    Please tell me this is right?

    <br />
a-(a^{\frac {1}{3}}b^\frac {2}{3})-b<br /> <br />
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  2. #2
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    Hello, Neenoon!

    You dropped a term . . . There are six multiplications.


    (a^{\frac{1}{3}} -b^{\frac{1}{3}})\,(a^{\frac{2}{3}}+a^{\frac{1}{3}}  b^{\frac{1}{3}} + b^{\frac{2}{3}})

    . . = \;(a^{\frac{1}{3}})(a^{\frac{2}{3}}) + (a^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) + (a^{\frac{1}{3}})(b^{\frac{2}{3}}) . - (b^{\frac{1}{3}})(a^{\frac{2}{3}}) - (b^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) - (b^{\frac{1}{3}})(b^{\frac{2}{3}})

    . . = \;a + \underbrace{a^{\frac{2}{3}}b^{\frac{1}{3}} + a^{\frac{1}{3}}b^{\frac{2}{3}} - a^{\frac{2}{3}}b^{\frac{1}{3}} - a^{\frac{1}{3}}b^{\frac{2}{3}}}_{\text{These cancel out}} - b
    . . = \;a - b

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  3. #3
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    Quote Originally Posted by Neenoon View Post
    (a^\frac {1}{3}-b^\frac {1}{3}) (a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3})

    (a^{\frac {1}{3}+\frac {2}{3}})+(a^{\frac {1}{3}+\frac {1}{3}}b^{\frac {1}{3}})-(b^{\frac {1}{3}}a^{\frac {2}{3}})-(b^{\frac {1}{3}+\frac {1}{3}}a^{\frac {1}{3}})-(b^{\frac {1}{3}+\frac {2}{3}})<br />

    <br />
(a^{\frac {3}{3}})+(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {1}{3}}b^\frac {2}{3})-(b^{\frac {3}{3}})<br />



    Please tell me this is right?

    <br />
a-(a^{\frac {1}{3}}b^\frac {2}{3})-b<br /> <br />
    Let's see...

    (a^\frac{1}{3}-b^\frac{1}{3})(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{  1}{3}+b^\frac{2}{3})<br />
=a^\frac{1}{3}(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{  1}{3}+b^\frac{2}{3})-b^\frac{1}{3}(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{1  }{3}+b^\frac{2}{3})
    =a+a^\frac{2}{3}b^\frac{1}{3}+a^\frac{1}{3}b^\frac  {2}{3}-a^\frac{2}{3}b^\frac{1}{3}-a^\frac{1}{3}b^\frac{2}{3}-b<br />
=a-b

    This is actually a case of the difference of two cubes rule...

    a^3-b^3=(a-b)(a^2+ab+b^2)
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  4. #4
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    I'll get back to math on monday, and check this out!

    I totally messed this one up... and I thought I was doing better.

    Thank you so much! It totally sounds right.

    ... but the actual answer is a-b, right?
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  5. #5
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    hi neoon
    it is a simple case of x^3 - y^3
    where x = a^1/3
    n y = b^1/3
    so ans will be a-b
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  6. #6
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    yes vishalgarg, totally simple ...
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