Question
Simplify the expression
a3ba5−2−2ba3
Evaluate
ba2−1÷(2a3b)−2
Multiply the terms
ba2−1÷2a3b−2
Divide the terms
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Evaluate
1÷2a3b
Multiply by the reciprocal
1×a3b2
Any expression multiplied by 1 remains the same
a3b2
ba2−a3b2−2
Reduce fractions to a common denominator
ba3a2×a3−a3b2−ba32ba3
Rewrite the expression
a3ba2×a3−a3b2−a3b2ba3
Write all numerators above the common denominator
a3ba2×a3−2−2ba3
Solution
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Evaluate
a2×a3
Use the product rule an×am=an+m to simplify the expression
a2+3
Add the numbers
a5
a3ba5−2−2ba3
Show Solution

Find the excluded values
b=0,a=0
Evaluate
ba2−1÷(2a3b)−2
To find the excluded values,set the denominators equal to 0
b=02a3b=0
Solve the equations
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Evaluate
2a3b=0
Multiply the terms
2a3b=0
Simplify
a3b=0
Separate the equation into 2 possible cases
a3=0b=0
The only way a power can be 0 is when the base equals 0
a=0b=0
b=0a=0b=0
Solution
b=0,a=0
Show Solution
