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#### karush

##### Well-known member

- Jan 31, 2012

- 3,062

$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$

OK this could get confusing quickly

but I don't think we want to square it first

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 3,062

$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$

OK this could get confusing quickly

but I don't think we want to square it first

- Jan 30, 2018

- 821

(The "2" in the denominator is already simple.)

$\frac{\left(ab^3-2a^{-4}b^3\right)^2}{4}$

Of course we can also factor out that $b^3$ that is in each term. Squaring it gives $b^6$.

$\frac{b^6(a- 2a^{-4})^2}{4}$

Now it is relatively easy to square $a- 2a^{-4}$.

$(a- 2a^{-4})^2= a^2- 2(a)(2a^{-4})+ (2a^{-4})^2$

$= a^2- 4a^{-3}+ 4a^{-8}$

$\frac{b^6(a^2-4a^{-3}+4a^{-8}}{4}$.

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- #3

- Jan 31, 2012

- 3,062

ok we still have negative exponents in the numerator??

- Jan 30, 2018

- 821

$\frac{b^2- 4a^{-3}+ 4a^{-8}}{4}= \frac{b^2- 4\frac{1}{a^{3}}+ 4\frac{1}{a^{8}}}{4}$

Now, to eliminate the "fractions in fractions", multiply both numerator and denominator by $a^8$

$\frac{a^8b^3- 4a^5+ 4}{4a^8}$

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- #5

- Jan 31, 2012

- 3,062

- Jan 30, 2018

- 821

That is exactly what I did!

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- #7

- Jan 31, 2012

- 3,062

yes but you created fractions over over a fraction $ \dfrac{a^{-3}}{4}=\dfrac{1}{4a^3}$That is exactly what I did!

anyway,, just being cranky

- Jan 30, 2018

- 821

No, [tex]a^{-3}[/tex] is not a fraction!

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- #9

- Jan 31, 2012

- 3,062

why not the exponent is positive

It is when it's written as $\displaystyle \begin{align*} \frac{1}{a^3} \end{align*}$...No, [tex]a^{-3}[/tex] is not a fraction!

- Jan 30, 2018

- 821

It is a matter of the difference in numerals, not numbers.

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- #12

- Jan 31, 2012

- 3,062

numerals???

Last edited:

Do you realise that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?

It is a matter of the difference in numerals, not numbers.

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- #14

- Jan 31, 2012

- 3,062

the bell rang and I left

no fancy stuff

- Jan 30, 2018

- 821

Yes, I am. They are two different numerals that repreent the same number. The term "fraction" has to do with the way a number is represented, it is not a property of the number itself.Do you realise that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?