Interesting Problem From Another Site

[soroban]

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

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

. . . . . . . $$=\;\underbrace{\left(a - \frac{1}{a}\right)}_{\text{This is 3}}\left(\underbrace{a^2 + \frac{1}{a^2}}_{\text{This is 11}} + 1\right)$$Therefore: $$\;a^3 - \frac{1}{a^3} \;=\;(3)(12) \;=\;36$$

 

[Fantini]

Perhaps the post margin was too small to contain it. (Giggle)