please help me!!!

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

please help me!!!

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Placentia, CA

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

Niceville, FL

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^{15}•1

-------- = ?

2b•a^{3}

-------- = ?

2b•a

Multiplication is commutative, so you can rearrange the denominator

a^{15}•1

-------- = ?

a^{3}•2b

-------- = ?

a

And you can combine like terms into separate fractions, like this

a^{15} 1

---- • ---- = ?

a^{3} 2b

---- • ---- = ?

a

Dividing numbers means subtracting the exponents of the denominator from the exponents of the numerator, so

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

and

a^{12} 1

---- • ---- = ?

1 2b

---- • ---- = ?

1 2b

Recombining results in

a^{12}

------ = ?

2b

2b

So the answer is

(a^{5}•b^{0})^{3 } a^{12}

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

(2b•a^{3}) 2b

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

(2b•a

Of 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!

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