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 January 8th, 2017, 05:49 AM #1 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Simplify the monster Ok so I woke up this morning to find this thing at the bottom of my bed. I've an exam tomorrow. This was from a previous paper. First bloody question on the exam. The poor critters. I had to take a pic because the thing was to hard to write Sorry I know it's a lil big. The picture I mean. Now then so I started by getting rid of them square roots. Ending up with just (c) on the numerator and b^2 on the bottom I was thinking to myself - Ok better distribute out the top part see what we've got. But that just ends up a proper mess. So I was thinking ok put them in brackets beside each other. (3a^1/2 x b^-1 x c) (3a^1/2 x b^-1 x c) Perhaps then I could use the powers on the bottom and just start subtracting away from the powers of the first term on the top...? Or maybe I need to multiply out the term using a common denominator. But then that ends up a mess as well. Can someone please push me in the general right direction. Thanks guys sorry I'm such a n00b. Last edited by Kevineamon; January 8th, 2017 at 06:34 AM.
 January 8th, 2017, 06:55 AM #2 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Had an idea. I was thinking using the power rule I could add up those powers in the two terms I've separated. Lets see hmmm (3a^1/2 x b^-1 x c) (3a^1/2 x b^-1 x c) 3a x 2b^-2 x c^2 Hmmm is that right? Can I just do that? Would avoid the distribution
 January 8th, 2017, 07:02 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,769 Thanks: 860 Got a headache as soon as I saw that ! Soooooo: http://www.wolframalpha.com/input/?i...4)*c%5E(-1%2F3)) I entered: x=(3*sqrt(a)*b^(-1)*sqrt(c^3))^2/(a^(-7)*sqrt(b^4)*c^(-1/3)) Thanks from Kevineamon Last edited by Denis; January 8th, 2017 at 07:07 AM.
 January 8th, 2017, 07:06 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 $\Large \dfrac{\left(3a^{1/2}b^{-1}\sqrt{c^3}\right)^2}{a^{-7}\sqrt{b^4}c^{-1/3}}=$ $\Large \dfrac{\left(3 a^{1/2} b^{-1}c^{3/2}\right)^2}{a^{-7}b^2 c^{-1/3}}=$ $\Large \dfrac{9 a b^{-2} c^3}{a^{-7}b^2 c^{-1/3}}=$ $\Large 9a^8 b^{-4}c^{10/3} =$ $\Large \dfrac{9 a^8 c^{10/3}}{b^4}$ Thanks from Kevineamon Last edited by romsek; January 8th, 2017 at 07:25 AM.
 January 8th, 2017, 07:12 AM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,052 Thanks: 431 There really is no way to avoid the work, but there really is not that much if you use PEMDAS. $\dfrac{ \left ( 3a^{(1/2)}b^{-1}\sqrt{c^3} \right )^2}{a^{-7}\sqrt{b^4}c^{-1/3}} =$ $\dfrac{9a b^{-2} c^3}{a^{-7} b^2 c^{-1/3}} =$ $\dfrac{9a * a^7 * c^3 * \sqrt[3]{c}}{b^2 * b^2} =$ $what?$ Thanks from Kevineamon
 January 8th, 2017, 07:14 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,769 Thanks: 860 Romsek, in denominator: c^(-1/3), not c^(1/3) Thanks from romsek and Kevineamon
 January 8th, 2017, 07:38 AM #7 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Thanks guys - interesting some of my ideas seemed to be correct, which shows my math brain is at least partially working. K let me stare at this thing for another while. Cheers guys you're the best
 January 8th, 2017, 08:01 AM #8 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Very interesting the way that $\displaystyle c^3/c^{-1/3} became$ $\displaystyle c^{10/3}$ Never knew that very interesting - thanks again guys
 January 8th, 2017, 09:32 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,769 Thanks: 860 Ya...tattoo this on your wrist: x^a / x^b = x^(a-b) ; 3^7 / 3^4 = 3^(7-4) = 3^3 Thanks from Kevineamon
 January 8th, 2017, 10:09 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,769 Thanks: 860 $\Large \dfrac{9 a^8 c^{10/3}}{b^4}$ Question: that of course implies that b<>0; will most teachers deduct a point or more if that's not specified as part of the answer?

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