$a^4+\frac{1}{a^4}=119$Then find $a^3+\frac{1}{a^3}$

I know $a^3+b^3=(a+b)(a^2+b^2-ab)$ and $a^4+b^4+a^2b^2=(a^2+b^2+ab)(a^2+b^2-ab)$

I try combining both but end up with equation in terms of a to the power 8, 7, 6 etc.

Printable View

- Jul 9th 2014, 04:55 AMNameIsHiddena^4+1/a^4=119. Then a^3+1/a^3
$a^4+\frac{1}{a^4}=119$Then find $a^3+\frac{1}{a^3}$

I know $a^3+b^3=(a+b)(a^2+b^2-ab)$ and $a^4+b^4+a^2b^2=(a^2+b^2+ab)(a^2+b^2-ab)$

I try combining both but end up with equation in terms of a to the power 8, 7, 6 etc. - Jul 9th 2014, 06:53 AMSlipEternalRe: a^4+1/a^4=119. Then a^3+1/a^3

Let . Then . So, and . This gives .

Then , so . Then or . Since any real number squared must be positive, it must be the first one.

So, or . This gives:

Since the square root of a positive number is positive, I put the first since and .

Hence, we have

Hence, . - Jul 9th 2014, 07:00 AMNameIsHiddenRe: a^4+1/a^4=119. Then a^3+1/a^3
- Jul 9th 2014, 08:43 AMSlipEternalRe: a^4+1/a^4=119. Then a^3+1/a^3

So, we have

Above, I showed .

So, , and - Jul 9th 2014, 09:11 AMNameIsHiddenRe: a^4+1/a^4=119. Then a^3+1/a^3
- Jul 9th 2014, 09:32 AMSlipEternalRe: a^4+1/a^4=119. Then a^3+1/a^3
I did all of the work for that in the post above.

- Jul 10th 2014, 04:14 AMNameIsHiddenRe: a^4+1/a^4=119. Then a^3+1/a^3
Thank you. I did not go through your first post properly.

- Jul 10th 2014, 05:56 AMJesterRe: a^4+1/a^4=119. Then a^3+1/a^3
Here's a way without actually solving for . First consider

so (already established)!

Then consider

so

Then consider

from which we obtain

.

Then substitute what we know giving

. - Jul 10th 2014, 08:10 AMJesterRe: a^4+1/a^4=119. Then a^3+1/a^3
There's a typo that I can't fix. The should be a .