My Math Forum Simplifying Algebraic Expressions using Exponent Rules

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 November 21st, 2013, 07:00 PM #1 Newbie   Joined: Nov 2013 Posts: 2 Thanks: 0 Simplifying Algebraic Expressions using Exponent Rules Hey, this is my first post - seems like a great site to visit to get help and to help others. I'm currently enrolled in the grade 11 University course named "Functions". I have a tutor, yet I still encounter some occasional speed bumps. Right now I'm connecting the exponent rules to simplify algebraic expressions. The power rules are simple enough, it's the co-efficients that stump me. So without further adieu, I'll provide the expression below, and I'd appreciate if somebody could show me the steps of completely simplifying it: (-2x^2)^3 (6x)^2 ------------------------- That dotted line is the division line. 2 (-3x^-1)^3 The answers in the back of the book got the answer as 16 ---- 3x Please illustrate how they reached this.
 November 21st, 2013, 07:06 PM #2 Newbie   Joined: Nov 2013 Posts: 2 Thanks: 0 Re: Simplifying Algebraic Expressions using Exponent Rules WAIT I MADE AN ERROR and I don't know how to edit. The actual expression that needs to be solved is: (-2x^-2)^3 (6x)^2 ----------------------- 2(-3x^-1)^3 Sorry! I hope nobody attempted the previous expresion!
 November 22nd, 2013, 05:18 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: Simplifying Algebraic Expressions using Exponent Rules $\frac{(-2x^{-2})^3(6x)^2}{2(-3x^{-1})^3}\,=\,\frac{-8x^{-6}\,\cdot\,36x^2}{2\,\cdot\,-27x^{-3}}\,=\,\frac{-8\,\cdot\,36}{2\,\cdot\,-27}\,\cdot\,\frac{x^{-6}\,\cdot\,x^2}{x^{-3}}\,=\,\frac{16}{3x}$

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