x+1=9x ^3+9x^2?

x+1 = 9x^3 + 9x^2

=> 9x^3 +9x^2 - x - 1 = 0

=>9x^2 (x+1) -1 (x+1) = 0

=>(x+1)(9x^2 -1) = 0

=>x = -1 or x^2 =1/9 =>x = +/- 1/3

ANSWER: x = -1 , +/-1/3

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

Therefore, x = -1, -1/3, +1/3

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

3x + 1 = 0
3x = -1
x = -1/3

3x - 1 = 0
3x = 1
x = 1/3

x + 1 = 0
x = -1

∴ x = ±1/3 , -1

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

x^2=1
x=-1
or x=1

(X+1) = 9X^2(X+1)

(X+1)/(X+1) = 9X^2

1= 9X^2

1/9 = X^2

X = +/- 1/3

f (x) = 9x³ + 9x² - x - 1
f (-1) = - 9 + 9 + 1 - 1 = 0
Thus x - (-1) is a factor of f (x)
ie x + 1 is a factor of f (x)
Find other factors by synthetic division :-
(-1)| 9 ____9____-1____-1
__ |_____- 9____ 0_____1
__ |9_____0____-1____ 0

f (x) = (x + 1)(9x² - 1) = 0
f (x) = (x + 1)(3x - 1)(3x + 1) = 0
x = - 1 , x = 1/3 , x = - 1/3

9x³ + 9x² - x - 1 = 0

(9x² - 1)(x + 1) = 0

(3x + 1)(3x - 1)(x + 1) = 0

x = 1/3, -1/3, or -1