What are the solutions to the equation y^3=y+24 ?

The solutions to the equation y^3=y+24 are y=3,y=-3/2+i(sqrt(23))/2 ,y=-3/2-i(sqrt(23))/2

Find the zeros of y^3=y+24

The zeros of y^3=y+24 are y=3,y=-3/2+i(sqrt(23))/2 ,y=-3/2-i(sqrt(23))/2

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y^{3} = y + 24

y = 3, y = - \frac{3}{2} + i \frac{23}{2}, y = - \frac{3}{2} - i \frac{23}{2}

Solution steps

y^{3} = y + 24

Move

24

to the left side

y^{3} - 24 = y

Move

y

to the left side

y^{3} - 24 - y = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

y^{3} - y - 24 = 0

Factor

y^{3} - y - 24 : (y - 3) (y^{2} + 3 y + 8)

(y - 3) (y^{2} + 3 y + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 3 = 0 or y^{2} + 3 y + 8 = 0

Solve

y - 3 = 0 : y = 3

Solve

y^{2} + 3 y + 8 = 0 : y = - \frac{3}{2} + i \frac{23}{2}, y = - \frac{3}{2} - i \frac{23}{2}

The solutions are

y = 3, y = - \frac{3}{2} + i \frac{23}{2}, y = - \frac{3}{2} - i \frac{23}{2}

d^{4} - 4 d^{3} + 8 d^{2} - 8 d + 4 = 0

(100) = 2 x^{100} ln (100) + 20

x^{3} + x^{2} + 3 x = - 4

x^{8} + x^{4} + 1 = 0

d^{3} + 3 d^{2} + 8 d - 8 = 0

What are the solutions to the equation y^3=y+24 ?
The solutions to the equation y^3=y+24 are y=3,y=-3/2+i(sqrt(23))/2 ,y=-3/2-i(sqrt(23))/2
Find the zeros of y^3=y+24
The zeros of y^3=y+24 are y=3,y=-3/2+i(sqrt(23))/2 ,y=-3/2-i(sqrt(23))/2