Can anyone solve this? 9x^4-10x^2+1=0?

f(1) = 9 - 10 + 1 = 0
Thus x - 1 is a factor
Obtain other factors by using synthetic division :-

1 | 9____0____-10____0____ 1
_ | _____9_____9____- 1____-1
_ | 9____9____- 1____- 1____ 0

(x - 1) (9x³ + 9x² - x - 1) = 0

- 1| 9____9____-1____-1
__|_____-9____ 0____ 1
__| 9____ 0____-1____0

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

(x-1)(x+1)(3x-1)(3x+1)=0
Therefore
x= 1, -1, 1/3, -1/3

9x^4 - 10x^2 + 1 = 0
(9x^2 - 1)(x^2 - 1) = 0
(3x + 1)(3x - 1)(x + 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 = 0
x = 1

∴ x = -1/3 , 1/3 , -1 , 1

9x^4-10x^2+1=0
let x^2 be y
then x^4 = y^2
it is now
9y^2-10y+1=0
9y^2-9y-y+1= 0
(9y^2-9y)-(y-1)=0
9y(y-1)-1(y-1) = 0
(y-1)(9y-1)=0
this implies
y = 1 or y = 1/9
since
x^2 = y
x^2 = 1
this implies x = 1,-1
also y = 1/9
since x^2 = y
x^2=1/9
this implies x = 1/3, x = -1/3

then
x = 1 , 1/3 , -1 , -1/3

Take y=x²
9x^4−10x^2+1=0
9y²−10y+1=0
9y²−9y−y+1=0
9y(y−1)−(y−1)=0
(y−1)(9y−1)=0
[Now resubstitute. I took 'y' so you will understand better. Its more better to do these steps without doing it.]
(x²−1)(9x²−1)=0
(x−1)(x+1)(3x−1)(3x+1)=0
x=1,−1,1/3,−1/3 is the required solution.
You can do this way also.
(x²−1)(9x²−1)=0
x²=1,1/9
x=1,−1,1/3,−1/3. Anyway its same.

This equation factors to
(9x^2 - 1)(x^2 - 1) = 0

This further factors to
(3x - 1)(3x + 1)(x + 1)(x - 1) = 0

Set all the factors to 0 and solve for x
3x - 1 = 0
x = 1/3

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

x + 1 = 0
x = -1

x - 1 = 0
x = 1

9x^4-10x^2+1=0

Let y=x^2

9y^2-10y+1=0
9y^2-9y-y+1=0

(y-1)(9y-1)=0

y=1 or y=1/9

re substituting y=x^2,

x^2=1 or x^2=1/9

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

Ans: x= -1 , 1 , -1/3 , 1/3

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