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x(x-1)(x+1)
do I multiply the brackets by eachother first, or each bracket by x?
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Multiplication is commutative, that is 6 x 7 = 7 x 6.
When you have three terms to be multiplied, first multiply the first and the second and the result by the third.It wouldn't make a difference if you multiply the second and the third and then multiply the result by the first.
x(x-1)(x+1) = (x²-x)(x+1) = x³ + x² - x² - x = x³ - x.
(x-1)(x+1)(x) = (x² - 1)(x) = x³ - x
As you can see, the results are the same in both the cases.
When you have a(x-y), you'd have to multiply both the terms inside the bracket by a, you get ax - ay.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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or you simply can see a formula (a-b)(a+b)=a² - b² ; as a result (x-1)(x+1)=x²-1²
so x(x-1)(x+1)=x(x²-1²)=x³-x ![]()
Last edited by Rose-Red (2006-01-05 21:54:54)
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Thankyou, I was drunk ![]()
Won't happen again.
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Maybe someone could simplify this?

again, I need the whole solution, because the answer is at the end of the book ![]()
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I'd start with x/x^2, firstly I'd "reciprocal" which basically means you take the value with the power (x^2) and move it to the top, when you do this you MUST change the sign of the power:
x / x^2 = x * x^-2
I'd then do the same with the next part of the sum, giving you b - ax * x^-3. Now all that is left is
(x * x^-2) + (b - ax * x^-3), since none of the values share similar powers, I think this is as far as you can go. (Don't take my word for it, I'm new to this too).
Last edited by rickyoswaldiow (2006-01-06 10:06:35)
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well the answer is 
I also tried to do that but in quite different way. Still I couldnt get the right answer. That's what I got
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Sounds like a job for Crazy Holmes ![]()
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I think the book's answer isn't correct. We can't eliminate a.
I got the same as you.
IPBLE: Increasing Performance By Lowering Expectations.
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yeah, maybe you're right. ![]()
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I've just read this "crazy holmes" thing.
Ha, ha, ha, that was VERY VERY... VERY
### ### ### ###
# # # # # # #
# ## # # # # # #
# # # # # # # #
### ### ### ###
IPBLE: Increasing Performance By Lowering Expectations.
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Actually it was BAD, but I'll forgive you if you promise that you will
NEVER
NEVER
NEVER
do this again
much
![]()
IPBLE: Increasing Performance By Lowering Expectations.
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appologise, grassy homes.
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Very funny. First krassi_holmz becomes crazy holmes, finally, grassy homes. I really hope krassi_holmz doesn't take offence at that. Maybe, rickyoswaldiow is just being funny. But a word of caution to rickyoswaldiow. People don't like their names (or usernames) spelt or pronouned wrongly. ![]()
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Yes, wildrocky, yes.
IPBLE: Increasing Performance By Lowering Expectations.
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No offence intended. I play online FPS games and my language in that would get me banned in a flash here. You **...
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I'm a bit late here, but I agree with everyone else.
I'd just scale up the left fraction by x to get a common denominator of x³, and then they can easily be added together.
Why did the vector cross the road?
It wanted to be normal.
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Yes, but the "a" cannot be removed.
IPBLE: Increasing Performance By Lowering Expectations.
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