please help me!!!

## (a5b0)3/(2ba3)=

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# 2 Answers

Correct, otherwise it would be:

(a5b0)3

---------

(2ba3)

multiplication by 0 in the numerator would make it:

0

----

6ba

and 0 divided by any number (other than 0) is 0:

0

I'm going to assume the equation you wrote is this:

(a

^{5}•b^{0})^{3}--------- = ?

(2b•a

^{3})The "cube" on the numerator's parentheses means

a5•b0 • a5•b0 • a5•b0

Since multiplying numbers means adding the exponents of the like terms, then

a

^{(5+5+5)}•b^{(0+0+0)}= a^{15}•b^{0}Thus you have

a

^{15}•b^{0}-------- = ?

2b•a

^{3}The zero exponent means "1", therefore

a

-------- = ?

2b•a

^{15}•1-------- = ?

2b•a

^{3}Multiplication is commutative, so you can rearrange the denominator

a

-------- = ?

a

^{15}•1-------- = ?

a

^{3}•2bAnd you can combine like terms into separate fractions, like this

a

---- • ---- = ?

a

^{15}1---- • ---- = ?

a

^{3}2bDividing numbers means subtracting the exponents of the denominator from the exponents of the numerator, so

a

^{(15-3)}= a^{12}and

a

---- • ---- = ?

1 2b

^{12}1---- • ---- = ?

1 2b

Recombining results in

a

^{12}------ = ?

2b

2b

So the answer is

(a

--------- = -----

(2b•a

^{5}•b^{0})^{3 }a^{12}--------- = -----

(2b•a

^{3}) 2bOf course, this is all assuming my interpretation of your problem is correct. Even if it's not, the principles shown here still hold.

Hope this helps!