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I know the answer is $243 \sqrt 3$ but in my maths book they got $(\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3)$ but then they only took the first $5$ out of the $11$ $(\sqrt 3)$s and then got $(3) (3) (3) (3) (3) (3) \sqrt 3$ and then got $3^5 \sqrt 3$

Why are they only using to the first $5$ $(\sqrt 3)$s? What happened to the other $6$ and how did the answer $243 \sqrt3$ arise? Thank you

GoodDeeds
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RTStriker
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    I feel like there might be a typo in your question: you say "then got $\underbrace{3\times\ldots\times3}_{6}\sqrt{3}$" but that would be one too many $3$s – Jam Sep 04 '18 at 17:00

3 Answers3

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What they are doing is grouping the $\sqrt 3$'s in pairs. There are $11\ \sqrt 3$s, so five pairs and one left over. Each pair becomes $\sqrt 3 \cdot \sqrt 3=3$, so they get five factors of $3$ and the left over $\sqrt 3$

Ross Millikan
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$$ \sqrt{3}^{11} = \sqrt{3} \cdot \sqrt{3}^{10} = \sqrt{3} \cdot \left(\sqrt{3}^2\right)^5 = \sqrt{3} \cdot 3^5 = 243\sqrt{3}. $$

MSDG
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Hint: $$3^5=243$$ and this is your answer.

Dr. Sonnhard Graubner
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