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Question Number 170389 by mr W last updated on 22/May/22

$$solveforx\in R$$$$x^3+\sqrt[3]{3-x}-3=0$$

Commented by Mastermind last updated on 22/May/22

$$Wecanactuallysolvethisbygraph$$$$method$$$$Okay,GraphDetails$$$$y=3-x^3$$$$y=\sqrt[3]{3-x}$$$$$$$$Root(\sqrt[3]{3},0)$$$$Domainx\in R$$$$Rangey\in R$$$$Verticalintercept(0,3)$$$$$$$$Root(3,0)$$$$Domainx\in R$$$$Rangey\in R$$$$Verticalintercept(0,\sqrt[3]{3})$$

Commented by mr W last updated on 22/May/22

$$whatisyouranswer?$$

Commented by Mastermind last updated on 22/May/22

$$youhavearrangeditagain$$

Commented by jasem1994hamoud last updated on 22/May/22

$$bywhatsap$$$$+79951154287$$

Commented by mr W last updated on 22/May/22

$$ijustputalltermsononeside.the$$$$equationisthesameasbefore.$$

Commented by Mastermind last updated on 22/May/22

$$youcanactuallyaddmetoo$$$$+2347068402491$$$$$$

Commented by Mastermind last updated on 22/May/22

$$Iunderstand$$

Answered by mr W last updated on 22/May/22

$$3-x^3=\sqrt[3]{3-x}$$$$f\left(x\right)=3-x^3\Rightarrow f^{-1}\left(x\right)=\sqrt[3]{3-x}$$$$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)$$$$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)=x$$$$3-x^3=xor\sqrt[3]{3-x}=x\left(thesame\right)$$$$x^3+x-3=0$$$$\Rightarrow x=\sqrt[3]{\sqrt{{\left(\frac{1}{3}\right)}^3+{(-\frac{3}{2})}^2}+\frac{3}{2}}-\sqrt[3]{\sqrt{{\left(\frac{1}{3}\right)}^3+{(-\frac{3}{2})}^2}-\frac{3}{2}}$$$$\Rightarrow x=\sqrt[3]{\frac{\sqrt{741}}{18}+\frac{3}{2}}-\sqrt[3]{\frac{\sqrt{741}}{18}-\frac{3}{2}}✓$$

Commented by Mastermind last updated on 22/May/22

$$Wow$$$${that}^{\prime }sgood$$

Commented by Rasheed.Sindhi last updated on 23/May/22

$$\mathbb{G}\mathbf{reat}\mathbb{S}\mathbf{ir}!$$

Commented by MJS_new last updated on 23/May/22

$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{also}\mathrm{has}\mathrm{got}\mathrm{a}\mathrm{pair}$$$$\mathrm{of}\mathrm{complex}\mathrm{solutions}.\mathrm{can}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{exact}$$$$\mathrm{values}?$$

Commented by mr W last updated on 23/May/22

$$i^{\prime }mnotsureifthecomplexrootsof$$$$x^3+x-3=0arealsothecomplexroots$$$$oftheoriginalequation.iguessyes,$$$$butnotsure.$$

Commented by MJS_new last updated on 23/May/22

$$\mathrm{no}.\mathrm{I}\mathrm{aporoximated}\mathrm{all}\mathrm{the}\mathrm{roots}$$$$\left(1\right)$$$$x^3+\sqrt[3]{3-x}-3=0$$$$x_1\approx 1.21341166$$$$x_{2,3}\approx -.597562988\pm .964934775\mathrm{i}$$$$\mathrm{these}\mathrm{roots}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{give}\mathrm{a}‘‘{\mathrm{nice}}^{″}\mathrm{polynome}:$$$$x^3-.0182856868x^2-1.56309342$$$$\left(2\right)$$$$x^3+x-3=0$$$$x_1\approx 1.21341166$$$$x_{2,3}\approx -.606705831\pm 1.45061225\mathrm{i}$$$$\left(3\right)$$$${[{(3-x)}^{1/3}=3-x^3]}^3$$$$\Rightarrow$$$$x^9-9x^6+27x^3-x-24=0$$$$(x^3+x-3)(x^6-x^4-6x^3+x^2+3x+8)=0$$$$x^6-4^4+6x^3+x^2+3x+8=0$$$$x_{1,2}\approx 1.54420759\pm .135231222\mathrm{i}$$$$x_{3,4}\approx -.597562988\pm .964934775\mathrm{i}$$$$x_{5,6}\approx -.946644598\pm 1.29938754\mathrm{i}$$

Commented by mr W last updated on 23/May/22

$$thanksforthisinterestingthing!$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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