Rozłóż na czynniki pierwsze: 81x^3-9x^2-9x+1 Pilne!! Proszę o pomoc.

9x^2(9x-1)-(9x-1)=0 (9x^2-1)(9x-1)=0 (3x-1)(3x+1)(9x-1)=0 x=1/3 lub x=-1/3 lub x=1/9

81x^3 -9x^2-9x+1= (81x^3 -9x^2)-(9x-1)= 9x^2(9x-1)-(9x-1)= (9x-1)(9x^2-1)= (9x-1)(3x-1)(3x+1)

81x³ - 9x²- 9x + 1 = 9x²(9x - 1) - (9x - 1) = (9x - 1)(9x² - 1) = = (9x - 1)(3x - 1)(3x + 1) = 81(x - ¹/₉)(x + ⅓)(x -⅓)