What are the solutions to the equation (x-1)^3-(x-2)^3=27 ?

The solutions to the equation (x-1)^3-(x-2)^3=27 are x=(9+sqrt(321))/6 ,x=(9-sqrt(321))/6

Find the zeros of (x-1)^3-(x-2)^3=27

The zeros of (x-1)^3-(x-2)^3=27 are x=(9+sqrt(321))/6 ,x=(9-sqrt(321))/6

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(x - 1)^{3} - (x - 2)^{3} = 27

x = \frac{9 + 321}{6}, x = \frac{9 - 321}{6}

Decimal Notation

x = 4.48607 \dots, x = - 1.48607 \dots

Solution steps

(x - 1)^{3} - (x - 2)^{3} = 27

Expand

(x - 1)^{3} - (x - 2)^{3} : 3 x^{2} - 9 x + 7

3 x^{2} - 9 x + 7 = 27

Move

27

to the left side

3 x^{2} - 9 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 20) = 321

x_{1, 2} = \frac{- ( - 9 ) \pm 321}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 321}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 321}{2 \cdot 3}

x = \frac{- ( - 9 ) + 321}{2 \cdot 3} : \frac{9 + 321}{6}

x = \frac{- ( - 9 ) - 321}{2 \cdot 3} : \frac{9 - 321}{6}

The solutions to the quadratic equation are:

x = \frac{9 + 321}{6}, x = \frac{9 - 321}{6}

x^{2} + x - 420 = 0

x^{2} + 136 x = 53.76

x^{2} = 1 - 6 x

4 x^{2} - 24 x + 16 = 0

3 x^{2} - 6 x - 36 = 0

What are the solutions to the equation (x-1)^3-(x-2)^3=27 ?
The solutions to the equation (x-1)^3-(x-2)^3=27 are x=(9+sqrt(321))/6 ,x=(9-sqrt(321))/6
Find the zeros of (x-1)^3-(x-2)^3=27
The zeros of (x-1)^3-(x-2)^3=27 are x=(9+sqrt(321))/6 ,x=(9-sqrt(321))/6