Simplifying Cube Roots in the DenominatorDate: 07/12/2001 at 08:39:18 From: pauline viray Subject: Simplifying cube roots in denominator I cannot simplify this: The cube root of 3 all over the (cube root of 2) minus the (cube root of 4). Thanks! Date: 07/12/2001 at 12:38:10 From: Doctor Jaffee Subject: Re: Simplifying cube roots in denominator Hi Pauline, I assume that what you did was multiply the numerator and denominator of your expression by cube root of 2 + cube root of 4. This process would work fine if you were trying to rationalize sqrt(2) - sqrt(4), but it doesn't work with cube roots. The reason your method works with square roots is that any expression in the form a^2 - b^2 can be factored to (a - b)(a + b), so if you have an expression in the form a - b and you multiply it by a + b, you end up with both terms being squared. In the problem above, however, you want both terms cubed. So, let's use the fact that a^3 - b^3 = (a - b)(a^2 + ab + b^2). In your problem a = cube root of 2 and b = cube root of 4. Multiply the numerator and denominator by a^2 + ab + b^2 and the resulting denominator will be cube root of 2^3 - cube root of 4^3 = -2. Give it a try and if you want to check your solution with me, write back. If you are having difficulties, let me know and show me what you have done so far, and I'll try to help you some more. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/