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!

Mark D.

Quantitative research SPECIALIST with academic/professional experience

New York, NY

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John P.

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Short Hills, NJ

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