how do you solve this equation?

(3x-6) (9x²-36x+36)=0
27(x-2) (x² - 4x + 4) = 0 (factor out "3" and the "9", 3x9 = 27)
27(x-2) (x - 2)² = 0
27(x-2)³ = 0
(x-2)³ = 0
x-2 = 0
x =2

(3x - 6)(9x² - 36x + 36) = 0
(3[x - 2])(3[3x² - 12x + 12]) = 0
(3[x - 2])(3[x - 2][3x - 6]) = 0
(3[x - 2])([3][3][x - 2]²) = 0
3³([x - 2]³) = 0
(3[x - 2])³ = 0
x - 2 = 0, x = 2

Answer: x = 2, the factors are: 3³(x - 2)³.

Proof:
(3[2] - 6)(9[2²] - 36[2] + 36) = 0
(6 - 6)(9[4] - 72 + 36) = 0
0(36 - 72 + 36) = 0
0(0) = 0

3x(9x^2-36x-36)-6(9x^2-36x-36)=0
27x^3-108x^2+108-54x^2+216x-216=0
(27x^3-162x^2+216x)=108
divide by 27
x^3-6x^2+8x=4
divide by x on the left side of the equa only
x^2-6x+8=4
(x+4)(x-2)=4
x+4 =4 or x+2=4
x=8 or x=2

(3x - 6)(9x^2 - 36x + 36) = 0
3 (x – 2) × 9(x^2 – 4x + 4) = 0
(x – 2)(x^2 – 4x + 4) = 0
(x – 2)(x – 2)² = 0
(x – 2)³ = 0
(x – 2) = 0
x = 2
∴ x = 2
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(3x-6) (9x^2-36x+36)=0
(x-2) (x^2-4x+4)=0
(x-2) (x-2)^2=0
(x-2)^3=0
the above equation is of degree three and hence it is supposed to have three roots. And these three roots as appearing are real and equal and that root is 2.
Hence the roots of given equation is said to be 2,2,2.

[6]
(3x-6)(9x^2-36x+36)=0
(3x-6)(3x-6)^2=0
3x-6=0
x=6/3=2 ans

To make things easier, let's take out all the numerical factors (3 and 9) first:

(3x-6)(9x^2-36x+36) = 0
(x-2)(x^2-4x+4) = 0

You can now see the factors to be
(x-2)(x-2)(x-2)

x=2

(3x - 6)(9x^2 - 36x + 36) = 0
(3x - 6)(3x - 6)(3x - 6) = 0

3x - 6 = 0
3x = 6
x = 6/3
x = 2

∴ x = 2

(3x-6) (9x^2-36x+36)=0
or, 9(3x-6)(x-2)^2=0
or,27(x-2)^3=0
or,x=2

actually there are three values of x all of which are equal to '2'
which is a case of multiple roots

This is the same as
(3x-6)(3x-6)(3x-6)
Does that help?