What are the solutions to the equation (x-4)^3=x^2(x-3) ?

The solutions to the equation (x-4)^3=x^2(x-3) are x= 8/3

Find the zeros of (x-4)^3=x^2(x-3)

The zeros of (x-4)^3=x^2(x-3) are x= 8/3

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(x - 4)^{3} = x^{2} (x - 3)

x = \frac{8}{3}

Decimal Notation

x = 2.66666 \dots

Solution steps

(x - 4)^{3} = x^{2} (x - 3)

Expand

(x - 4)^{3} : x^{3} - 12 x^{2} + 48 x - 64

Expand

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

x^{3} - 12 x^{2} + 48 x - 64 = x^{3} - 3 x^{2}

Move

3 x^{2}

to the left side

x^{3} - 9 x^{2} + 48 x - 64 = x^{3}

Move

x^{3}

to the left side

- 9 x^{2} + 48 x - 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 9 ) ( - 64 )}{2 ( - 9 )}

4 8^{2} - 4 (- 9) (- 64) = 0

x_{1, 2} = \frac{- 48 \pm 0}{2 ( - 9 )}

x = \frac{- 48}{2 ( - 9 )}

\frac{- 48}{2 ( - 9 )} = \frac{8}{3}

x = \frac{8}{3}

The solution to the quadratic equation is:

x = \frac{8}{3}

x^{2} + 2 x + 7 = 2 x - 5

- 4.905 t^{2} + 16 - 7 = 0

\frac{4}{3} x^{2} - \frac{2}{5} = 0

x = 2 (x - 1)^{2} + 8

2 x^{2} + 6 = 8 x

What are the solutions to the equation (x-4)^3=x^2(x-3) ?
The solutions to the equation (x-4)^3=x^2(x-3) are x= 8/3
Find the zeros of (x-4)^3=x^2(x-3)
The zeros of (x-4)^3=x^2(x-3) are x= 8/3