Rationalizing Denominators with Cube RootsDate: 06/26/2008 at 16:38:42 From: Lea Subject: Rationalizing Denominators Dear Dr. Math, How do you rationalize this denominator: 1 ------------------------------ cubrt(6) + cubrt (9) + cubrt(4) I find it hard to use sums of cubes. I grouped cubrt(6) and cubrt(9) so can use sums of cubes but the problem became more complicated. Is there a technique to rationalize this expression? Date: 06/28/2008 at 11:41:34 From: Doctor Ali Subject: Re: Rationalizing Denominators Hi Lea! Thanks for writing to Dr. Math. So we have, 1 --------------------------- __ __ __ 3 / 3 / 3 / \/ 6 + \/ 9 + \/ 4 We have cube roots in the denominator, so we have to multiply something by top and bottom to make cubes to cancel out the cube roots. Do you know this formula? (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = a^3 + b^3 + c^3 - 3abc We can let a = cbrt(6) b = cbrt(9) c = cbrt(4) So, we have (a + b + c) in the denominator. We can multiply top and bottom by (a^2 + b^2 + c^2 - ab - ac - bc) to get rid of the cube roots. In other words, we should multiply top and bottom by, __ __ __ __ __ __ 3 / 3 / 3 / 3 / 3 / 3 / S = \/ 36 + \/ 81 + \/ 16 - \/ 54 - \/ 24 - \/ 36 Does that make sense? The numerator will be S itself. The denominator will be __ __ __ __________ 3 / 3 3 / 3 3 / 3 3 / \/ 6 + \/ 9 + \/ 4 - \/ 6 x 9 x 4 Which is, 6 + 9 + 4 + 6 = 25 a rational number! Does that make sense? You might also want to take a look at: Rationalizing Denominators http://www.mathforum.org/library/drmath/view/53052.html This is another example of rationalizing denominators with odd roots in the denominator. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
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