Challenge Problem #3

Soroban · **Posted:** Tue, 12 Aug 2003 20:22:03 UTC

Challenge Problem #3

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

bugzpodder · **Joined:** Sun, 4 May 2003 16:04:19 UTC **Posts:** 2906

WooHoo!

let the equation be:
a-b=2
(a-b)^2=a^2-2ab+b^2=4

a^3-b^3=56=(a-b)(a^2+ab+b^2)=(2)(4+3ab)
so 56=2(4+3ab) and
ab=8

(a-b)^2+4ab=(a+b)^2=4+32=36

if a+b=6, and a=4,b=2
6x+28=4^3, x=6
if a+b=-6
a=-2,b=-4
6x+28=-8
x=-6

Soroban · **Posted:** Tue, 12 Aug 2003 23:09:56 UTC

Nice going Bugz!
I was sure that this was your kind of problem.

The book has a more devious solution.
It's quite a learning experience.
It follows immediately after the wavy line.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Theorem: If a + b + c = 0

then

(A proof is given at the end.)

We have: $\sqrt[3]{6x + 28} - \sqrt[3]{6x - 28} - 2 = 0$

Let $a = \sqrt{6x + 28},\; b = -\sqrt[3]{6x - 28},\; c = -2$

From the theorem, we have:

$(\sqrt[3]{6x + 28})^3 - (\sqrt[3]{6x - 28})^3 + (-2)^3 = 3\sqrt[3]{6x + 28}\cdot \sqrt[3]{6x - 28}(2)$

$(6x + 28) - (6x - 28) - 8 = 6\sqrt[3]{36x^2 - 784}$

$48 = 6\sqrt[3]{36x^2 - 784}$

$\sqrt[3]{36x^2 - 784} = 8$

36x^2 - 784 = 512