[SOLVED] Check this algebric operation

**Neenoon** · August 15th 2008, 05:29 PM

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

$(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}}) $

$ (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}}) $

Please tell me this is right?

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

**Soroban** · August 15th 2008, 06:21 PM

Hello, Neenoon!

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

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

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

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

**Prove It** · August 15th 2008, 06:21 PM

Originally Posted by Neenoon

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

$(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}}) $

$ (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}}) $

Please tell me this is right?

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

Let's see...

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

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

$a^3-b^3=(a-b)(a^2+ab+b^2)$

**Neenoon** · August 16th 2008, 05:57 AM

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?

**vishalgarg** · August 16th 2008, 06:08 AM

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

**Neenoon** · August 19th 2008, 04:08 PM

yes vishalgarg, totally simple ...

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