SOLUTION: the cube root of 8x to the 3rd-1 = 2x-1 i already tried cubing both sides of the equation to get rid of the radical then then i was left with: 8x to the 3rd -1= 8x to the 3rd

Algebra ->  Square-cubic-other-roots -> SOLUTION: the cube root of 8x to the 3rd-1 = 2x-1 i already tried cubing both sides of the equation to get rid of the radical then then i was left with: 8x to the 3rd -1= 8x to the 3rd      Log On


   

Question 693710: the cube root of 8x to the 3rd-1 = 2x-1
i already tried cubing both sides of the equation to get rid of the radical then then i was left with:
8x to the 3rd -1= 8x to the 3rd -12x to the 2nd +6x-1
this is where i got stuck what did i do wrong above?
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
root%283%2C8x%5E3-1%29+=+2x-1
...
root%283%2C8x%5E3-1%29%5E3+=+%282x-1%29%5E3
8x^3-1 = (2x-1)^3
...
(2x-1)^3 = (2x-1)(4x^2-4x+1)
8x^3-8x^2+2x-4x^2+4x-1
8x^3-12x^2+6x-1
...
8x^3 - 1 = 8x^3 - 12x^2 + 6x - 1
0 = 8x^3 - 8x^3 - 12x^2 + 6x - 1 + 1
0 = -12x^2 + 6x
0 = -6x(2x-1)
0 = 2x-1
2x = 1
x = 1%2F2
and
x = 0
...
You did not make any errors, you just needed to keep going!
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