Cubing a Negative Fraction

[Jason76]

$\displaystyle - \dfrac{3}{2} * - \dfrac{3}{2} * - \dfrac{3}{2}$

It should come out to:

$\displaystyle - \dfrac{27}{8}$


However, a video said the answer was the same as $\displaystyle \dfrac{8}{27}$

Is there some rule at play here?

 

[MarkFL]

You are right and the video is wrong. The cube of a negative is negative. Think of the graph of:

$\displaystyle y=x^3$

 

[lookagain]

$\displaystyle - \dfrac{3}{2} * - \dfrac{3}{2} * - \dfrac{3}{2}$

It should come out to:

$\displaystyle - \dfrac{27}{8}$


However, a video said the answer was the same as $\displaystyle \dfrac{8}{27}$

Is there some rule at play here?

Jason76,

would you check the problem on the video to see if it is actually

$\displaystyle \bigg(\dfrac{3}{2}\bigg)^{-3} \ ?$

 

[Jason76]

Jason76,

would you check the problem on the video to see if it is actually

$\displaystyle \bigg(\dfrac{3}{2}\bigg)^{-3} \ ?$

That's what it is. So, therefore the answer would be different than if it was only $\displaystyle \dfrac{3}{2}^{-3}$. Right?

 

[JeffM]

That's what it is. So, therefore the answer would be different than if it was only $\displaystyle \dfrac{1}{3}^{-3}$. Right?

I doubt you asked the question that you actually wanted to ask.

Obviously $\displaystyle \dfrac{1}{3}^{-3} \ne \left(\dfrac{3}{2}\right)^{-3}.$

Some laws of exponents

$\displaystyle \left(\dfrac{a}{b}\right)^c = \dfrac{a^c}{b^c}.$

$\displaystyle d \ne 0 \implies d^{-c} = \dfrac{1}{d^c}.$

From the two laws above we can deduce:

$\displaystyle e\ne 0 \implies \left(\dfrac{e}{f}\right)^{-c} = \dfrac{1}{\left(\dfrac{e}{f}\right)^c} = \dfrac{1}{\dfrac{e^c}{f^c}} = \dfrac{1}{1} * \dfrac{f^c}{e^c} = \dfrac{f^c}{e^c} = \left(\dfrac{f}{e}\right)^c.$

 

[Jason76]

I doubt you asked the question that you actually wanted to ask.

Obviously $\displaystyle \dfrac{1}{3}^{-3} \ne \left(\dfrac{3}{2}\right)^{-3}.$

Some laws of exponents

$\displaystyle \left(\dfrac{a}{b}\right)^c = \dfrac{a^c}{b^c}.$

$\displaystyle d \ne 0 \implies d^{-c} = \dfrac{1}{d^c}.$

From the two laws above we can deduce:

$\displaystyle e\ne 0 \implies \left(\dfrac{e}{f}\right)^{-c} = \dfrac{1}{\left(\dfrac{e}{f}\right)^c} = \dfrac{1}{\dfrac{e^c}{f^c}} = \dfrac{1}{1} * \dfrac{f^c}{e^c} = \dfrac{f^c}{e^c} = \left(\dfrac{f}{e}\right)^c.$

Sorry I meant 3 over 2 to the negative 1/3 power.

Right. Here is an analogy:

Pretend c is -1.

$\displaystyle (\dfrac{a}{b})^{-1} = \cfrac{1}{\cfrac{a}{b}} = \cfrac{\cfrac{b}{a}}{1} = \dfrac{b}{a}$

 

[Deleted member 4993]

That's what it is. So, therefore the answer would be different than if it was only $\displaystyle \dfrac{3}{2}^{-3}$. Right?

That is again a totally different problem.


$\displaystyle \dfrac{3}{2}^{-3} \ = \ \dfrac{1}{27} * \dfrac{1}{2} \ = \ \dfrac{1}{54}$

 

[pappus]

Sorry I meant 3 over 2 to the negative 1/3 power.

...

Do I understand that correctly that you want to evaluate

$\displaystyle \displaystyle{\left( \frac32 \right)^{-\frac13} = \left( \frac23 \right)^{\frac13} = \frac13 \cdot \sqrt[3]{18}}$

 

[JeffM]

Jason

Let's start with another law of exponents.

$\displaystyle integer\ q > 0 \implies p^{1/q} = \sqrt[q]{p}\ BY\ DEFINITION.$

So $\displaystyle integer\ q >0 \implies p^{r/q} = \left(p^r\right)^{1/q} = \sqrt[q]{p^r}.$

Therefore

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\left(\dfrac{3}{2}\right)^{(1/3)}}.$ Negative exponent law. OK so far?

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{\dfrac{3}{2}}}.$ Fractional exponent law. Still with me?

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{1.5}}.$ Simplification.

Actually it is easier to simplify as the first step but I wanted to concentrate on the laws of exponents.

EDIT: Pappua and I arrive at the same answer, just expressed differently

$\displaystyle \left(\dfrac{1}{2} * \sqrt[3]{18}\right)^3 = \dfrac{1}{27} * 18 = \dfrac{2}{3}.$

$\displaystyle \left(\dfrac{1}{\sqrt[3]{1.5}}\right)^3 = \dfrac{1}{1.5} = \dfrac{1}{1.5} * \dfrac{2}{2} = \dfrac{2}{3}.$

He used a different law of exponents.

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{3^{-(1/3)}}{2^{-(1/3)}}.$ He started here.

So $\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{\dfrac{1}{3^{(1/3)}}}{\dfrac{1}{2^{(1/3)}}} = \dfrac{1}{\sqrt[3]{3}} * \dfrac{\sqrt[3]{2}}{1} = \dfrac{\sqrt[3]{2}}{\sqrt[3]{3}} =\dfrac{2^{(1/3)}}{3^{(1/3)}} = \left(\dfrac{2}{3}\right)^{(1/3)}.$ Now he rationalized.

$\displaystyle \left(\dfrac{2}{3}\right)^{-(1/3)} = \left(\dfrac{2}{3}\right)^{(1/3)} = \left(\dfrac{2 * 9}{3 * 9}\right)^{(1/3)} = \left(\dfrac{18}{27}\right)^{(1/3)} = \dfrac{18^{(1/3)}}{27^{(1/3)}} = \dfrac{\sqrt[3]{18}}{3}.$

 

[Jason76]

Jason

Let's start with another law of exponents.

$\displaystyle integer\ q > 0 \implies p^{1/q} = \sqrt[q]{p}\ BY\ DEFINITION.$

So $\displaystyle integer\ q >0 \implies p^{r/q} = \left(p^r\right)^{1/q} = \sqrt[q]{p^r}.$

Therefore

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\left(\dfrac{3}{2}\right)^{(1/3)}}.$ Negative exponent law. OK so far?

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{\dfrac{3}{2}}}.$ Fractional exponent law. Still with me?

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{1.5}}.$ Simplification.

Actually it is easier to simplify as the first step but I wanted to concentrate on the laws of exponents.

EDIT: Pappua and I arrive at the same answer, just expressed differently

$\displaystyle \left(\dfrac{1}{2} * \sqrt[3]{18}\right)^3 = \dfrac{1}{27} * 18 = \dfrac{2}{3}.$

$\displaystyle \left(\dfrac{1}{\sqrt[3]{1.5}}\right)^3 = \dfrac{1}{1.5} = \dfrac{1}{1.5} * \dfrac{2}{2} = \dfrac{2}{3}.$

He used a different law of exponents.

$\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{3^{-(1/3)}}{2^{-(1/3)}}.$ He started here.

So $\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{\dfrac{1}{3^{(1/3)}}}{\dfrac{1}{2^{(1/3)}}} = \dfrac{1}{\sqrt[3]{3}} * \dfrac{\sqrt[3]{2}}{1} = \dfrac{\sqrt[3]{2}}{\sqrt[3]{3}} =\dfrac{2^{(1/3)}}{3^{(1/3)}} = \left(\dfrac{2}{3}\right)^{(1/3)}.$ Now he rationalized.

$\displaystyle \left(\dfrac{2}{3}\right)^{-(1/3)} = \left(\dfrac{2}{3}\right)^{(1/3)} = \left(\dfrac{2 * 9}{3 * 9}\right)^{(1/3)} = \left(\dfrac{18}{27}\right)^{(1/3)} = \dfrac{18^{(1/3)}}{27^{(1/3)}} = \dfrac{\sqrt[3]{18}}{3}.$

Evaluate the power:

Note: I tried to put the parenthesis, but I don't know the La Tex. So just assume there is parenthesis between exponent and fraction.

$\displaystyle \dfrac{3}{2}^{-3}$

Rewrite the fraction as 1 over a positive exponent (negative exponent law).

$\displaystyle \cfrac{1}{\cfrac{3}{2}^{3}}$

Evalutate the exponent

$\displaystyle \cfrac{1}{\cfrac{27}{8}}$

1 over a fraction = the reciprocal of a fraction over 1.

$\displaystyle \cfrac{\cfrac{8}{27}}{1} = \dfrac{8}{27}$ Answer

This makes it more simple.

 

[pka]

Evaluate the power:
Note: I tried to put the parenthesis, but I don't know the La Tex. So just assume there is parenthesis between exponent and fraction.
This makes the job much simpler.

Evaluate the power:
$\displaystyle \cfrac{\cfrac{2}{3}^{3/1}}{1} = \dfrac{2}{3}^{3/1} = \dfrac{8}{27}$ Answer
This makes the job much simpler.

Note the use of \left( \right)
\left(\dfrac{3}{2}\right)^{-3}[/tex] gives $\displaystyle \left(\dfrac{3}{2}\right)^{-3}$

To make is simpler $\displaystyle \left(\dfrac{3}{2}\right)^{-3}= \left(\dfrac{2}{3}\right)^{3}$.

That is, do the inversion first of all.

 

[Jason76]

Do I understand that correctly that you want to evaluate

$\displaystyle \displaystyle{\left( \frac32 \right)^{-\frac13} = \left( \frac23 \right)^{\frac13} = \frac13 \cdot \sqrt[3]{18}}$

Example

Note: Assume parenthesis are between the exponent and fraction. Problem with understanding La tex.

$\displaystyle \dfrac{3}{2}^{-1/3} = \cfrac{1}{\cfrac{3}{2}^{1/3}}$

Next we would have 1 over the cubed root of 3 over 2 (because it can be re-written in cubed form). I don't know how to go further than that, cause there doesn't seem to be any exact cubes.

1 over cubed root of 3 over 2 - Answer