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Some hard alg2 question

0
333
3

Find the exact result of:

$$(\sqrt{3}-2)^{2017}*(\sqrt{3}+2)^{2018}$$

edit: my bad, bases aren't the same

Dec 22, 2017
edited by Guest  Dec 22, 2017

#1
+2

OK! I thought there was more to it than that!.

The exact result is =-(2 + sqrt(3)) - According to "Mathematica 11 Home Edition"

Note: The steps it uses are soooooooo Loooooooooooong, it is rediculous!! There has to be a simpler way of doing it. Maybe somebody else can come up with a short answer.

Dec 22, 2017
edited by Guest  Dec 22, 2017
#2
+7348
+2

$$(\sqrt3-2)^{2017}(\sqrt3+2)^{2018} \\~\\ =\,(\sqrt3-2)^{2017}(\sqrt3+2)^{2017}(\sqrt 3+2) \\~\\ =\,[\,(\sqrt3-2)(\sqrt3+2)\,]^{2017}\,(\sqrt3+2) \\~\\ =\,[\,3-4\,]^{2017}\,(\sqrt3+2) \\~\\ =\,[-1]^{2017}\,(\sqrt3+2) \\~\\ =\,(-1)(\sqrt3+2) \\~\\ =\,-\sqrt3-2$$

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Dec 22, 2017
#3
+94619
+2

Nice, hectiictar!!!!

[ This one was a little easier after the edit......LOL!!!  ]

Dec 22, 2017