# [SOLVED] Check this algebric operation

• Aug 15th 2008, 05:29 PM
Neenoon
[SOLVED] Check this algebric operation
$\displaystyle (a^\frac {1}{3}-b^\frac {1}{3})$$\displaystyle (a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3}) \displaystyle (a^{\frac {1}{3}+\frac {2}{3}})+(a^{\frac {1}{3}+\frac {1}{3}}b^{\frac {1}{3}})-(b^{\frac {1}{3}}a^{\frac {2}{3}})-(b^{\frac {1}{3}+\frac {1}{3}}a^{\frac {1}{3}})-(b^{\frac {1}{3}+\frac {2}{3}}) \displaystyle (a^{\frac {3}{3}})+(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {1}{3}}b^\frac {2}{3})-(b^{\frac {3}{3}}) Please tell me this is right? (Doh) \displaystyle a-(a^{\frac {1}{3}}b^\frac {2}{3})-b • Aug 15th 2008, 06:21 PM Soroban Hello, Neenoon! You dropped a term . . . There are six multiplications. \displaystyle (a^{\frac{1}{3}} -b^{\frac{1}{3}})\,(a^{\frac{2}{3}}+a^{\frac{1}{3}} b^{\frac{1}{3}} + b^{\frac{2}{3}}) . . \displaystyle = \;(a^{\frac{1}{3}})(a^{\frac{2}{3}}) + (a^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) + (a^{\frac{1}{3}})(b^{\frac{2}{3}}) .\displaystyle - (b^{\frac{1}{3}})(a^{\frac{2}{3}}) - (b^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) - (b^{\frac{1}{3}})(b^{\frac{2}{3}}) . . \displaystyle = \;a + \underbrace{a^{\frac{2}{3}}b^{\frac{1}{3}} + a^{\frac{1}{3}}b^{\frac{2}{3}} - a^{\frac{2}{3}}b^{\frac{1}{3}} - a^{\frac{1}{3}}b^{\frac{2}{3}}}_{\text{These cancel out}} - b . . \displaystyle = \;a - b • Aug 15th 2008, 06:21 PM Prove It Quote: Originally Posted by Neenoon \displaystyle (a^\frac {1}{3}-b^\frac {1}{3})$$\displaystyle (a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3})$

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

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

Please tell me this is right? (Doh)

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

Let's see...

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

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

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$
• Aug 16th 2008, 05:57 AM
Neenoon
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. (Doh) (Angry) (Crying)

Thank you so much! It totally sounds right.

... but the actual answer is a-b, right?
• Aug 16th 2008, 06:08 AM
vishalgarg
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
• Aug 19th 2008, 04:08 PM
Neenoon
yes vishalgarg, totally simple ... (Surprised) (Headbang) (Sweating) (Sweating) (Sweating) (Sweating)