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This question is pissing me off

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This question is pissing me off

by jeffboshine » Sat Apr 03, 2010 1:54 pm
If abc = b^3 , which of the following must be true?


1. ac = b^2
2. b = 0
3. ac = 1


I chose 1.

The right answer was none.

Can someone help? Everything I know about algebra tells me that if abc=b^3, then ac=b^2.

The explanation states that for B=0, item 1 does not hold. We don't have enough information to make that assertion. If B = 0, then ac=0.

Any help on this one. It's ticking me off.
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by kevincanspain » Sat Apr 03, 2010 2:47 pm
jeffboshine wrote:If abc = b^3 , which of the following must be true?


1. ac = b^2
2. b = 0
3. ac = 1


I chose 1.

The right answer was none.

Can someone help? Everything I know about algebra tells me that if abc=b^3, then ac=b^2.

The explanation states that for B=0, item 1 does not hold. We don't have enough information to make that assertion. If B = 0, then ac=0.

Any help on this one. It's ticking me off.

I see what you are trying to do: you are dividing both sides by b, which you can do as long as you know that b is not equal to 0

You could do the following: abc - b^3 = 0
b(ac -b^2)= 0 so b=0 OR ac=b^2
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by eaakbari » Sat Apr 03, 2010 10:45 pm
All three options mentioned are all possibilities, and will satisfy the equation if substituted . But there is not enough information to conclude from barely
abc = b^3
Remember you can only cancel b out if b is not equal to 0. If you do cut it you have an option of b = 0

ac = b^2 or b= 0

Now ac =1 will also satisfy but thats not the only solution b=6, a=2 c=18 will also satisfy , so will b = 0

Hence none as its ambiguous
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