Help with (a^5 b^-2 c^-1)^-3 * (a^-5 b^-2 c)^-3 and

[mathnoob]

Could you please help me with the following two questions, as I am unsure how to simply them fully. Here is what I have so far.

Question 1: (a^5 b^-2 c^-1)^-3 * (a^-5 b^-2 c)^-3

a^-15 b^6 c^3 * a^15 b^-6 c^-3

After this I am unsure of my next step...

Question 2: (x^3 y^-3 z^-3)^-2 / (y^5)^-2 ÷ (y^3 z^5)^-1

= (x^-6 y^6 z^6) / (y^-10) ÷ (y^-3 z^-5)

= again this is as far as I have gotten

If someone could show me steps they use I would appreciate it.
Thanks people

 

[fasteddie65]

Re: Help with two algebra questions.

For Question 1, simply add the exponents of like variables. For example, a^-15 • a^15 = a^0 = 1

For Question 2, simplify the second part by expanding the outside exponents, then dividing.
(y^5)^-2 ÷ (y^3 z^5)^-1 = y^-10 ÷ (y^-3 z^-5) = y^-7 z^5

 

[mathnoob]

ok thanks...so for the first question...my answer would be 3?

as for the second....I still don't see what you did there . where did the x^-6 go?

 

[stapel]

owhere did the x^-6 go?

I think that's a typo...?

To learn how the exponent rules work, try here. For worked examples of how to simplify with exponents, try here.

\(\displaystyle \mbox{1) }\, (a^5 b^{-2} c^{-1})^{-3}\, \times\, (a^{-5} b^{-2} c)^{-3}\)

You were told to start with \(\displaystyle (a^{-15} b^{6} c^{3})\, \times \,(a^15) b^{6} c^{-3})\) and then simplify from there. How did you arrive at a numerical value of "3"? Please show your steps.

I will have to guess your meaning for the second exercise, since the two "divided by" signs aren't clear as to grouping.

\(\displaystyle \mbox{2) }\, \frac{(x^3 y^{-3} z^{-3})^{-2}}{\left(\frac{(y^5)^{-2}}{(y^3 z^5)^{-1}}\right)}\)

A good first step would be to restate the above as:

. . . . .\(\displaystyle \frac{\left(\frac{y^6 z^6}{x^6}\right)}{\left(\frac{y^3 z^5}{y^{10}}\right)}\)

Then convert the "divided by a fraction" to "multipled by the reciprocal", and simplify.

If you get stuck, please reply showing how far you have gotten. Thank you!

 

[Denis]

...so for the first question...my answer would be 3?

No. You have a slight error here:
> Question 1: (a^5 b^-2 c^-1)^-3 * (a^-5 b^-2 c)^-3
> a^-15 b^6 c^3 * a^15 b^-6 c^-3
Your b^-6 should be b^6 ; OK?
So you'll end up with b^12

 

[mathnoob]

No. You have a slight error here:
> Question 1: (a^5 b^-2 c^-1)^-3 * (a^-5 b^-2 c)^-3
> a^-15 b^6 c^3 * a^15 b^-6 c^-3
Your b^-6 should be b^6 ; OK?
So you'll end up with b^12


Sorry for the typo..the equation is (a^5 b^-2 c^-1)^-3 * (a^-5 b^2 c)^-3

which would make it b^-6 right?

 

[mathnoob]

I will have to guess your meaning for the second exercise, since the two "divided by" signs aren't clear as to grouping.

(x^3 y^-3 z^-3)^-2 / (y^5)^-2 ÷ (y^3 z^5)^-1
this part is a fraction divided by the second part

 

[Denis]

Sorry for the typo..the equation is (a^5 b^-2 c^-1)^-3 * (a^-5 b^2 c)^-3
which would make it b^-6 right?

Yep. So 1*1*1 = 1

 

[Denis]

(x^3 y^-3 z^-3)^-2 / (y^5)^-2 ÷ (y^3 z^5)^-1
this part is a fraction divided by the second part

WHICH part?
Is it (x^3 y^-3 z^-3)^-2 / [(y^5)^-2 / (y^3 z^5)^-1] or [(x^3 y^-3 z^-3)^-2 / (y^5)^-2] / (y^3 z^5)^-1 ??

And WHY use 2 different symbols to represent "divide"?

 

[mathnoob]

Question 2: (x^3 y^-3 z^-3)^-2 / (y^5)^-2 ÷ (y^3 z^5)^-1

the first divide side is supposed to represent a fraction...sorry ...so it is (x^3 y^-3 z^-3)^-2 / (y^5)^-2 which is the first part divided by (y^3 z^5)^-1
sorry for the confusion

 

[Denis]

Question 2: (x^3 y^-3 z^-3)^-2 / (y^5)^-2 ÷ (y^3 z^5)^-1
= (x^-6 y^6 z^6) / (y^-10) ÷ (y^-3 z^-5)
= again this is as far as I have gotten

So now established:
[(x^-6 y^6 z^6) / (y^-10)] / (y^-3 z^-5)

RULE: (a / b) / c = a / (bc)
So above becomes:
x^-6 y^6 z^6 / [(y^-10)(y^-3 z^-5)]

= x^-6 y^6 z^6 / (y^-13 z^-5)

= y^19 z^11 / x^6 : OK?

Regarding your note:
> the first divide side is supposed to represent a fraction...

Remember that a fraction IS a division; as example:
fraction 3/2 means 3 divided by 2 : kapish?

Getting technically silly:
3 means 3 divided by 1 :wink:

 

[mathnoob]

very descriptive..thanks for all the help!