How do you solve #root[ 3] { 8x ^ { 3} - 1} = 2x - 1#?

1 Answer
Noah G
Nov 29, 2016

#x= 0 and 1/2#.

Explanation:

#(root(3)(8x^3 - 1))^3 = (2x- 1)^3#

#8x^3 - 1 = (2x- 1)(2x- 1)(2x - 1)#

#8x^3 - 1 = (4x^2 - 4x + 1)(2x- 1)#

#8x^3 - 1 = 8x^3 - 8x^2 + 2x - 4x^2 + 4x - 1#

#0 = 8x^3 - 8x^3 + 1 - 8x^2 + 2x - 4x^2 + 4x - 1#

#0 = -12x^2 + 6x#

#0 = 6x(-2x + 1)#

#x = 0 and 1/2#

Hopefully this helps!

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