# formula question

• Apr 27th 2008, 08:29 AM
yuriythebest
formula question
• Apr 27th 2008, 08:35 AM
flyingsquirrel
Hi

No, it's wrong, unless * is the symbol you use for the sum and + the one you use for the product... :D
• Apr 27th 2008, 08:36 AM
Moo
Hello,

Not it isn't.

$a^{{\color{red}n}*k}$ (I suppose it's n, not a) is equal to $(a^n)^k=(a^k)^n$

The only thing you can do is to factor by $a^k$, if $k :

$a^n+a^k=a^k \left(a^{n-k}+1 \right)$
• Apr 27th 2008, 08:36 AM
Mathstud28
Quote:

Originally Posted by yuriythebest

no.....the only manipulation I could see would be this $a^n+a^k=a^n(1+a^{k-m})$

and to test your hypothesis let a=2 and k=1 and n=2
• Apr 27th 2008, 08:48 AM
yuriythebest
wow thanks for the lighting-quick responses. But it seems I am still doing something terribly wrong:

http://img516.imageshack.us/img516/4...tion001ah6.jpg
• Apr 27th 2008, 08:50 AM
Moo
I'd make a substitution :

$u=(x^3+8)^{1/4}$

Then the equation is now uČ+u=6 :)
• Apr 27th 2008, 09:54 AM
yuriythebest
ok I tried it that way but it seems I am going around in circles...

http://img135.imageshack.us/img135/1...tion002gt4.jpg
• Apr 27th 2008, 10:03 AM
flyingsquirrel
No, you're not "going around in circles". Once you have found the two solutions $u_1,\,u_2$ of $u^2+u-6=0$, you can solve for $x$ : $\sqrt[4]{x^3+8}=u_1 \Rightarrow x^3+8=u_1^4$ and so on...

Be careful about $\Delta=b^2-4ac=1^2-4\times 1\times (-6)\neq 40$
• Apr 27th 2008, 10:26 AM
yuriythebest
ok I tried that and got 2 negative roots:

http://img340.imageshack.us/img340/3...tion003vq8.jpg
• Apr 27th 2008, 10:32 AM
flyingsquirrel
OK you got the idea, now, read again the post #8. (especially the last sentence "Be careful...") :D
• Apr 27th 2008, 11:20 AM
yuriythebest
thank you flyingsquirrel! answer iz correct!
• Apr 27th 2008, 10:56 PM
earboth
Quote:

Originally Posted by yuriythebest
thank you flyingsquirrel! answer iz correct!

I'm referring to post #7: I don't understand where you've got the results from:

$u^2+u-6=0~\implies~u = -3~\vee~u=2$

And therefore

$\sqrt[4]{x^3+8} = -3$ that means there doesn't exist a real x to satisfy this equation

or

$\sqrt[4]{x^3+8} = 2~\implies~x^3+8=16~\implies~x^3=8~\implies~\boxed {x=2}$
• Apr 28th 2008, 04:04 AM
yuriythebest
yeah I understood that. cause of my spelling I mistook one of my u's for a 4 and therefore the incorrect answer. With your help I redid it and everything is correct.