What are the solutions to the equation 1*y+y^3=2 ?

The solutions to the equation 1*y+y^3=2 are y=1,y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2

Find the zeros of 1*y+y^3=2

The zeros of 1*y+y^3=2 are y=1,y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2

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1 \cdot y + y^{3} = 2

y = 1, y = - \frac{1}{2} + i \frac{7}{2}, y = - \frac{1}{2} - i \frac{7}{2}

Solution steps

1 \cdot y + y^{3} = 2

Refine

y + y^{3} = 2

Move

2

to the left side

y + y^{3} - 2 = 0

Write in the standard form

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

y^{3} + y - 2 = 0

Factor

y^{3} + y - 2 : (y - 1) (y^{2} + y + 2)

(y - 1) (y^{2} + y + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 1 = 0 or y^{2} + y + 2 = 0

Solve

y - 1 = 0 : y = 1

Solve

y^{2} + y + 2 = 0 : y = - \frac{1}{2} + i \frac{7}{2}, y = - \frac{1}{2} - i \frac{7}{2}

The solutions are

y = 1, y = - \frac{1}{2} + i \frac{7}{2}, y = - \frac{1}{2} - i \frac{7}{2}

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

(x - 3)^{3} - 1 = 26

x^{3} + 4 x^{2} - 8 x = 32

9 x^{2} (x + 5) - 4 (x + 5) = 0

2 x^{3} - 5 x^{2} + x + 2 = 0

What are the solutions to the equation 1*y+y^3=2 ?
The solutions to the equation 1*y+y^3=2 are y=1,y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2
Find the zeros of 1*y+y^3=2
The zeros of 1*y+y^3=2 are y=1,y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2