This is some wacky thing, Soroban, or does it supposed to have a viable answer.
If so, I got -6 and 6
Hello, OReilly!
Well done!
The "Quickie" solution uses a clever theorem,
. . but is not much shorter than your solution.
Theorem: .If a + b + c .= .0, then: .a³ + b³ + c³ .= . 3abc **
. . . . . . . . . _______ . . ______
We have: .³√6x + 28 - ³√6x - 28 - 8 .= .0
-. . . . . . . . . . .↑ . . . . . . . ↑ . w . ↑
. . . . . . . . . . . a . . . . . . . b . o . c
The thereom gives us: . . . . . . . . ._________________
. . (6x + 28) - (6x - 28) - 8 .= .3·³√(6x + 28)(6x - 28)(8)
. . . . . . . . . . . . . . . . . . . . . . . . _________
. . . . . . . . . . . . . . . . .48 .= .6·³√36x² - 784
Then: .36x² - 784 .= .512 . → . x² = 36 . → . x = ±6
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** . Proof
We have: .a + b .= .-c
Cube both sides: .(a + b)³ .= .(-c)³
Expand: .aª + 3a²b + 3ab² + b³ .= .-c³
Then: .a³ + b³ + c³ .= .-3a²b - 3ab²
. . . . . a³ + b³ + c³ .= .-3ab(a + b)
Since a + b = -c, we have: .a³ + b³ + c³ .= .-3ab(-c)
Therefore: . a³ + b³ + c³ .= .3abc