# Math Help - Quickie #2

1. ## Quickie #2

Solve: . $\sqrt[3]{6x + 28} - \sqrt[3]{6x - 28} \:=\:2$

2. This is some wacky thing, Soroban, or does it supposed to have a viable answer.

If so, I got -6 and 6

3. Originally Posted by Soroban

Solve: . $\sqrt[3]{6x + 28} - \sqrt[3]{6x - 28} \:=\:2$

$\begin{array}{l}
\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}} = 2 \\
\left( {\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}} \right)^3 = 2^3 \\
\end{array}
$

$6x + 28 - 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}) - 6x + 28 = 8$
$- 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(\sqrt[3]{{6x + 28}} - \sqrt[3]{{6x - 28}}) = - 48$
$- 3\sqrt[3]{{6x + 28}}\sqrt[3]{{6x - 28}}(2) = - 48$
$\begin{array}{l}
3\sqrt[3]{{(6x + 28)(6x - 28)}} = 24 \\
\sqrt[3]{{(6x + 28)(6x - 28)}} = 8 \\
(6x + 28)(6x - 28) = 512 \\
36x^2 - 784 = 512 \\
36x^2 = 1296 \\
x^2 = 36 \\
x = \pm 6 \\
\end{array}
$

4. Soroban has a clever trick to solve this.

5. Originally Posted by ThePerfectHacker
I think the trick is that we need to check the solutions. Because those the necessary but not sufficient conditions.
Thus,
x\not = -6
Why x=-6 isn't solution?

$\sqrt[3]{{6( - 6) + 28}} - \sqrt[3]{{6( - 6) - 28}} = \sqrt[3]{{ - 8}} - \sqrt[3]{{ - 64}} = - 2 - ( - 4) = 2$

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