x-3-x-2-x-1-3-Find-the-value-of-x-Help-me-please-

Question Number 215177 by Ismoiljon_008 last updated on 30/Dec/24

x^{3} + x^{2} + x = - \frac{1}{3}

F i n d t h e v a l u e o f x!

H e l p m e, p l e a s e

Answered by mnjuly1970 last updated on 30/Dec/24

3 x^{3} + 3 x^{2} + 3 x + 1 = 0

2 x^{3} + {(x + 1)}^{3} = 0

(\sqrt[3]{2} x + x + 1) ({\sqrt[3]{4} x^{2} - x^{2} \sqrt[3]{2} - \sqrt[3]{2} x + x^{2} + 2 x + 1} > 0) = 0

- x (\sqrt[3]{2} + 1) = 1 \Rightarrow x = \frac{- 1}{1 + \sqrt[3]{2}}

Answered by Ghisom last updated on 31/Dec/24

x^{3} + x^{2} + x + \frac{1}{3} = 0

x = t - \frac{1}{3}

t^{3} + \frac{2}{3} t + \frac{2}{27} = 0

{Cardano}^{'} s Formula gives

t_{1} = - \frac{2^{2 / 3}}{3} + \frac{2^{1 / 3}}{3}

\Rightarrow x_{1} = - \frac{2^{2 / 3} - 2^{1 / 3} + 1}{3}

t_{2} = - \frac{2^{2 / 3}}{3} ω + \frac{2^{1 / 3}}{3} ω^{2}

\Rightarrow x_{2} = \frac{2^{2 / 3} - 2^{1 / 3} - 2}{6} + \frac{(2^{2 / 3} + 2^{1 / 3}) \sqrt{3}}{6} i

t_{3} = - \frac{2^{2 / 3}}{3} ω^{2} + \frac{2^{1 / 3}}{3} ω

\Rightarrow x_{3} = \frac{2^{2 / 3} - 2^{1 / 3} - 2}{6} - \frac{(2^{2 / 3} + 2^{1 / 3}) \sqrt{3}}{6} i

Commented by MathematicalUser2357 last updated on 04/Jan/25

\frac{2^{2 / 3} - 2^{1 / 3} - 2}{6} - \frac{(2^{2 / 3} + 2^{1 / 3}) \times \sqrt{3}}{6} \times i

(- . 278753 - . 821951 i)