What are the solutions to the equation e=(x+1)(x^2-x+1) ?

The solutions to the equation e=(x+1)(x^2-x+1) are x\approx 1.19774…

Find the zeros of e=(x+1)(x^2-x+1)

The zeros of e=(x+1)(x^2-x+1) are x\approx 1.19774…

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

x \approx 1.19774 \dots

Solution steps

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

Expand

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

e = x^{3} + 1

Switch sides

x^{3} + 1 = e

Move

e

to the left side

x^{3} + 1 - e = 0

Find one solution for

x^{3} - 1.71828 \dots = 0

using Newton-Raphson:

x \approx 1.19774 \dots

Apply long division:

\frac{x ^{3} + 1 - e}{x - 1.19774 \dots} = x^{2} + 1.19774 \dots x + 1.43459 \dots

x^{2} + 1.19774 \dots x + 1.43459 \dots \approx 0

Find one solution for

x^{2} + 1.19774 \dots x + 1.43459 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 1.19774 \dots

x, (x - 1)^{3} = y^{3}

- 2 x^{4} + 9 = 3 x - 1

x^{4} - 3 x - 20 = 0

x^{23} x - 4 + 4 x = 10 + x^{2}

x^{3} - 4 x + x - 2 = 0

What are the solutions to the equation e=(x+1)(x^2-x+1) ?
The solutions to the equation e=(x+1)(x^2-x+1) are x\approx 1.19774…
Find the zeros of e=(x+1)(x^2-x+1)
The zeros of e=(x+1)(x^2-x+1) are x\approx 1.19774…