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$\dfrac{(8^3)(-16)^5}{4(-2)^8}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{4\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

is it equal to... ? $$ \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} $$

I am bit confused, How can I handle this problem ?

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Franck, noting the pattern of your last few questions, I offer a tiny tip: factor anything that can be factored. It helps a great deal in seeing what cancels. –  J. M. Sep 12 '11 at 17:41
    
Thanks ! Finally.. I understood. :) –  Franck Sep 12 '11 at 17:45

2 Answers 2

Hint: Just take everything to powers of 2 using the laws of exponents

$$\frac{8^3(-16)^5}{4(-2)^8}=\frac{-(2^3)^3(2^4)^5}{2^2\cdot2^8}=?$$

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I don't like this. It's really hard to get students to cancel before multiplying. Often they see $\dfrac{20\cdot19\cdot18\cdot17\cdot16\cdot15}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$ and they start by multiplying out the numerator and the denominator! –  Michael Hardy Sep 12 '11 at 18:38
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@Michael Hardy: to me it depends upon the problem. For this one, especially without a calculator, all those powers of 2 scream at me and I would like an answer of -(2^n) as I know those (not quite this high). I can do $5\cdot 4+3 \cdot 3-2-8$ in my head. –  Ross Millikan Sep 12 '11 at 18:43

When multiplying fractions, cancel BEFORE multiplying.

$$ \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} $$

Wherever you see $\dfrac{8}{2}$, put $4$.

Wherever you see $\dfrac{-16}{2}$, put $-8$.

Then you have $$ 4\cdot4\cdot4\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8). $$

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