Question

Hi, I need a little help with solving this question for c:

(2c-1)^3=4c^2-1

I did this already:
(2c-1)^3-(4c^2-1)=0
(2c-1)^3-(2c+1)(2c-1)=0

But I need help with the next steps... Thank you!

Answer 1

(2c-1)^3-(2c+1)(2c-1)=0
good so far, now take out a common factor
(2c-1)[(2c-1)^2 - (2c+1)] = 0 expand and simplify to
(2c-1)(4c^2 - 6c) = 0
(2c-1)(c)(4c-6) = 0
c = 1/2 or c = 0 or c = 3/2