1.1.21 simplify rational expression

[karush]

simplify
$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$
OK this could get confusing quickly
but I don't think we want to square it first

 

[HOI]

Yes, it can be simplified by cancelling "a" before squaring.
(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 $a- 2a^{-4}$.
$(a- 2a^{-4})^2= a^2- 2(a)(2a^{-4})+ (2a^{-4})^2$
$= a^2- 4a^{-3}+ 4a^{-8}$

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

 

[karush]

ok we still have negative exponents in the numerator??

 

[HOI]

Is that a problem? Different people have different ideas of what "simple" is. If you don't want negative exponents, remember what a negative exponent means:
$\frac{b^2- 4a^{-3}+ 4a^{-8}}{4}= \frac{b^2- 4\frac{1}{a^{3}}+ 4\frac{1}{a^{8}}}{4}$

Now, to eliminate the "fractions in fractions", multiply both numerator and denominator by $a^8$
$\frac{a^8b^3- 4a^5+ 4}{4a^8}$

 

[karush]

well the few examples I saw they brought the terms with negative exponents in the numerator to the denumerator

 

[HOI]

That is exactly what I did!

 

[karush]

That is exactly what I did!

yes but you created fractions over over a fraction $ \dfrac{a^{-3}}{4}=\dfrac{1}{4a^3}$

anyway,, just being cranky

 

[HOI]

No, a^{-3} is not a fraction!

 

[karush]

why not the exponent is positive

 

[Prove It]

No, a^{-3} is not a fraction!

It is when it's written as $\displaystyle \begin{align*} \frac{1}{a^3} \end{align*}$...

 

[HOI]

Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

 

[karush]

numerals?

 

[Prove It]

Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

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

 

[karush]

they just said move the terms with negative exponents in the numerator to the denominator
the bell rang and I left
no fancy stuff

 

[HOI]

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

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.