Hi guys. I've been wondering if you could possibly help me. I don't know what to do with that 27 on the second term.
Last edited by Alienis Back; Sep 29th 2008 at 06:48 AM. Reason: Forgot to attach the problem
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Think of it as $\displaystyle (4)^3 + (3a)^3$
That's really useful...
Originally Posted by Alienis Back That's really useful... Yes it is. Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a... $\displaystyle 4^3 + x^3 = (4+x)(16-4x-x^2)$
Originally Posted by Chop Suey Yes it is. Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a...[/tex] That is a good idea. Originally Posted by Chop Suey $\displaystyle 4^3 + x^3 = (4+x)(16-4x-x^2)$ But I think here there is a mistake. Look at the formula for the sum of cubes above.
Originally Posted by Alienis Back That is a good idea. But I think here there is a mistake. Look at the formula for the sum of cubes above. Yes, you are right. I made a typo there. it should be +x^2 rather than -x^2.
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