That would work if it were 3b instead of b^3. As it is, they're different numbers. In this case because abc=b^3, you can't take away a b from one side and make it b^2.

Does that make sense?

What's the OA?

Agreed on both questions- looks like you got the answers switched.

If b=0 that won't work. Example:

abc=b^3

a=2
b=0
c=3

abc=0, b^3=0
but
ac=6, b^2=0

so that eliminates I.

if:

a=2
b=4
c=8

abc=64, b^3=64

which eliminates II and III also

So, I'd go with A. OA?

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Re: how much does A have?

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