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# (a5b0)3/(2ba3)=

### 2 Answers by Expert Tutors

Brian B. | Begining Math and Algebra Tutor in Placentia area.Begining Math and Algebra Tutor in Place...
4.9 4.9 (15 lesson ratings) (15)
0
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:

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Nathan C. | Math and Science (K through College)Math and Science (K through College)
5.0 5.0 (23 lesson ratings) (23)
0
I'm going to assume the equation you wrote is this:

(a5•b0)3
---------   =  ?
(2b•a3)

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)  =  a15•b0

Thus you have

a15•b0
--------  = ?
2b•a3

The zero exponent means "1", therefore

a15•1
-------- = ?
2b•a3

Multiplication is commutative, so you can rearrange the denominator

a15•1
-------- = ?
a3•2b

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

a15     1
---- • ---- = ?
a3      2b

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

a(15-3) = a12

and

a12      1
---- • ---- = ?
1      2b

Recombining results in

a12
------ = ?
2b

(a5•b0)3     a12
--------- =  -----
(2b•a3)       2b

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!