SOLUTION: `Pls help out: {{{ x^(2/3) -5x^(1/3) + 6=0 }}}

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Question 947406: `Pls help out: +x%5E%282%2F3%29+-5x%5E%281%2F3%29+++++%2B+++++6=0+++
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^(2/3) - 5x^(1/3) + 6 = 0
----
Let w = x^(1/3)
----
Rewrite the quadratic::
w^2 - 5w + 6 = 0
---
Factor::
(w-6)(w+1) = 0
--------
w = 6 or w = -1
------------------
Solve for "x"::
If w = 6, x^(1/3) = 6 ; x = 6^3 = 216
----
If w = -1, x^(1//3) = -1 ; x = (-1)^3 = -1
=====================
Cheers,
Stan H.
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Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

`Pls help out: +x%5E%282%2F3%29+-5x%5E%281%2F3%29+++++%2B+++++6=0+++


highlight%28highlight%28x%5E%282%2F3%29+-+5x%5E%281%2F3%29+%2B+6+=+0%29%29

Let highlight%28a+=+x%5E%281%2F3%29%29

Then: highlight%28highlight%28x%5E%282%2F3%29+-+5x%5E%281%2F3%29+%2B+6+=+0%29%29 becomes: a%5E2+-+5a+%2B+6+=+0

(a - 3)(a - 2) = 0

a = 3       OR       a = 2



a = 3

Since a was substituted for highlight%28highlight%28x%5E%281%2F3%29%29%29, then we can say that:

highlight%28highlight%283+=+x%5E%281%2F3%29%29%29

3%5E3+=+%28x%5E%281%2F3%29%29%5E3 ------- Cubing both sides

highlight_green%2827+=+x%29



a = 2

Since a was substituted for highlight%28highlight%28x%5E%281%2F3%29%29%29, then we can say that:

highlight%28highlight%282+=+x%5E%281%2F3%29%29%29

2%5E3+=+%28x%5E%281%2F3%29%29%5E3 ------- Cubing both sides

highlight_green%288+=+x%29