MHB Interesting Problem From Another Site

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An intriguing mathematical problem from MHF involves finding the value of a^3 - 1/a^3 given that a^4 + 1/a^4 = 119. The proposed solution involves manipulating the equation by adding 2 to both sides, leading to the expression (a^2 + 1/a^2)^2 = 121, which simplifies to a^2 + 1/a^2 = 11. Further calculations reveal that a - 1/a = 3, allowing the derivation of a^3 - 1/a^3 using the formula a^3 - 1/a^3 = (a - 1/a)(a^2 + 1 + 1/a^2). The final result is a^3 - 1/a^3 = 36, highlighting the elegance of the solution despite posting difficulties on the original site.
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There is an interesting problem at MHF.
I think I have an elegant solution for it,
but the site won't let me post my reply.
(It's only on this one problem. Go figure!)
Anyway, I just have to share my solution.

a^4+\frac{1}{a^4} \:=\:119.\;\;\text{Find }\,a^3-\frac{1}{a^3}
Add 2 to both sides: \:a^4 + 2 + \frac{1}{a^4} \:=\:121 \quad\Rightarrow\quad \left(a^2 + \frac{1}{a^2}\right)^2 \:=\: 121

Hence: \;a^2 + \frac{1}{a^2} \:=\:11Subtract 2 from both sides: \:a^2 - 2 + \frac{1}{a^2} \;=\;9 \quad\Rightarrow\quad \left(a - \frac{1}{a}\right)^2 \:=\:9

Hence: \; a - \frac{1}{a} \:=\:3a^3 - \frac{1}{a^3} \;=\;\left(a-\frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right)

. . . . . . . =\;\underbrace{\left(a - \frac{1}{a}\right)}_{\text{This is 3}}\left(\underbrace{a^2 + \frac{1}{a^2}}_{\text{This is 11}} + 1\right)Therefore: \;a^3 - \frac{1}{a^3} \;=\;(3)(12) \;=\;36
 
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Perhaps the post margin was too small to contain it. (Giggle)
 
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