What are the solutions to the equation (3x-2)^3=(2-3x)^2 ?

The solutions to the equation (3x-2)^3=(2-3x)^2 are x=1,x= 2/3

Find the zeros of (3x-2)^3=(2-3x)^2

The zeros of (3x-2)^3=(2-3x)^2 are x=1,x= 2/3

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(3 x - 2)^{3} = (2 - 3 x)^{2}

x = 1, x = \frac{2}{3}

Decimal Notation

x = 1, x = 0.66666 \dots

Solution steps

(3 x - 2)^{3} = (2 - 3 x)^{2}

Expand

(3 x - 2)^{3} : 27 x^{3} - 54 x^{2} + 36 x - 8

Expand

(2 - 3 x)^{2} : 4 - 12 x + 9 x^{2}

27 x^{3} - 54 x^{2} + 36 x - 8 = 4 - 12 x + 9 x^{2}

Move

9 x^{2}

to the left side

27 x^{3} - 63 x^{2} + 36 x - 8 = 4 - 12 x

Move

12 x

to the left side

27 x^{3} - 63 x^{2} + 48 x - 8 = 4

Move

4

to the left side

27 x^{3} - 63 x^{2} + 48 x - 12 = 0

Factor

27 x^{3} - 63 x^{2} + 48 x - 12 : 3 (x - 1) (3 x - 2)^{2}

3 (x - 1) (3 x - 2)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 3 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x - 2 = 0 : x = \frac{2}{3}

The solutions are

x = 1, x = \frac{2}{3}

x, a = (2 x^{5} y^{3})^{3}

a^{3} + 12 a - 6.5 = 0

8 (1.2 - x^{2})^{2} = 0.5 x^{2}

76 = \frac{1}{2} (h (h^{2} + 3))

x^{2} = 8 x^{15}

What are the solutions to the equation (3x-2)^3=(2-3x)^2 ?
The solutions to the equation (3x-2)^3=(2-3x)^2 are x=1,x= 2/3
Find the zeros of (3x-2)^3=(2-3x)^2
The zeros of (3x-2)^3=(2-3x)^2 are x=1,x= 2/3