If (x+\(\frac{1}{2}\))³ + (3x+\(\frac{2}{3}\))³ + (x-\(\frac{1}{6}\))³ = \(\frac{1}{12}\)(9x+2)(2x+1)(6x-1) 

find the value of x².

\(\frac{1}{25}\)

If a³ + b³ + c³ = 3abc

then a + b + c = 0

x+\(\frac{1}{2}\)+3x+\(\frac{2}{3}\)+x-\(\frac{1}{6}\) = 0

5x+\(\frac{1}{2}\)+\(\frac{2}{3}\)-\(\frac{1}{6}\)=0

x=\(\frac{-1}{5}\)

x²=(\(\frac{-1}{5}\))² = \(\frac{1}{25}\)