How do you factor $x^{3} + x = 3$ $x^3 + x = 3$ ?

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How do you factor $x^{3} + x = 3$ $x^3 + x = 3$ ?

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Answer 1 · 2016-05-30T14:28:59

$x^{3} + x - 3 = (x - x_{1}) (x - x_{2}) (x - x_{3})$ $x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)$

where:

$x_{1} = \frac{1}{3} ⎛ ⎝ \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}} ⎞ ⎠$ $x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))$

etc.

Explanation:

See:
How do you find all the real and complex roots of $f (x) = x^{3} + x - 3$ $f(x) = x^3+x-3$ ?

There we find zeros:

$x_{1} = \frac{1}{3} ⎛ ⎝ \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}} ⎞ ⎠$ $x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))$

$x_{2} = \frac{1}{3} ⎛ ⎝ ω \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + ω^{2} \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}} ⎞ ⎠$ $x_2 = 1/3(omega root(3)((81+3sqrt(741))/2)+omega^2 root(3)((81-3sqrt(741))/2))$

$x_{3} = \frac{1}{3} ⎛ ⎝ ω^{2} \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + ω \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}} ⎞ ⎠$ $x_3 = 1/3(omega^2 root(3)((81+3sqrt(741))/2)+omega root(3)((81-3sqrt(741))/2))$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$

Subtract $3$ $3$ from both sides of the given equation to get:

$x^{3} + x - 3 = 0$ $x^3+x-3 = 0$ .

Then $x^{3} + x - 3 = (x - x_{1}) (x - x_{2}) (x - x_{3})$ $x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)$

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How do you factor $x^{3} + x = 3$ $x^3 + x = 3$ ?

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