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Question Number 156992 by MathSh last updated on 18/Oct/21

Commented by MJS_new last updated on 19/Oct/21

$$\mathrm{I}\mathrm{get}$$$$x=1\vee x\approx -.428605913602$$$$\mathrm{no}\mathrm{other}\mathrm{solution}\in \mathbb{C}$$

Answered by Rasheed.Sindhi last updated on 18/Oct/21

$$\frac{3}{\sqrt[3]{\mathrm{x}+1}}+\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^3+1}}=2\sqrt[3]{4}$$$${(\frac{3}{\sqrt[3]{\mathrm{x}+1}}+\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^3+1}}=2\sqrt[3]{4})}^3$$$$▸\frac{27}{\mathrm{x}+1}+\frac{{\mathrm{x}}^3}{{\mathrm{x}}^3+1}$$$$+3\left(\frac{3}{\sqrt[3]{\mathrm{x}+1}}\right)\left(\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^3+1}}\right)(\frac{3}{\sqrt[3]{\mathrm{x}+1}}+\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^3+1}})=8\left(4\right)$$$$▸\frac{27}{\mathrm{x}+1}+\frac{{\mathrm{x}}^3}{{\mathrm{x}}^3+1}+3\left(\frac{3}{\sqrt[3]{\mathrm{x}+1}}\right)\left(\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^3+1}}\right)\left(2\sqrt[3]{4}\right)=8\left(4\right)$$$$▸\frac{27}{\mathrm{x}+1}+\frac{{\mathrm{x}}^3}{{\mathrm{x}}^3+1}+\left(\frac{9\mathrm{x}}{\sqrt[3]{({\mathrm{x}}^3+1)(\mathrm{x}+1)}}\right)\left(2\sqrt[3]{4}\right)=8\left(4\right)$$$$▸\frac{18\sqrt[3]{4}\mathrm{x}}{\sqrt[3]{({\mathrm{x}}^3+1)(\mathrm{x}+1)}}=32-\frac{27}{\mathrm{x}+1}-\frac{{\mathrm{x}}^3}{{\mathrm{x}}^3+1}$$$$=\frac{32(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)-27({\mathrm{x}}^2-\mathrm{x}+1)+{\mathrm{x}}^3}{(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)}$$$$=\frac{32({\mathrm{x}}^3+1)-27({\mathrm{x}}^2-\mathrm{x}+1)+{\mathrm{x}}^3}{(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)}$$$$=\frac{{33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5}{{\mathrm{x}}^3+1}$$$$▸{\left(\frac{18\sqrt[3]{4}\mathrm{x}}{\sqrt[3]{({\mathrm{x}}^3+1)(\mathrm{x}+1)}}\right)}^3={\left(\frac{{33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5}{{\mathrm{x}}^3+1}\right)}^3$$$$\frac{5832\left(4\right){\mathrm{x}}^3}{({\mathrm{x}}^3+1)(\mathrm{x}+1)}=\frac{{({33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5)}^3}{{({\mathrm{x}}^3+1)}^3}$$$$\frac{5832\left(4\right){\mathrm{x}}^3}{\mathrm{x}+1}=\frac{{({33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5)}^3}{{({\mathrm{x}}^3+1)}^2}$$$$\frac{5832\left(4\right){\mathrm{x}}^3}{\mathrm{x}+1}=\frac{{({33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5)}^3}{{(\mathrm{x}+1)}^2{({\mathrm{x}}^2-\mathrm{x}+1)}^2}$$$$5832\left(4\right){\mathrm{x}}^3=\frac{{({33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5)}^3}{(\mathrm{x}+1){({\mathrm{x}}^2-\mathrm{x}+1)}^2}$$$$▸5832\left(4\right){\mathrm{x}}^3(\mathrm{x}+1){({\mathrm{x}}^2-\mathrm{x}+1)}^2$$$$-{({33\mathrm{x}}^3-{27\mathrm{x}}^2+27\mathrm{x}+5)}^3=0$$$$$$$$\mathrm{x}=-0.423807\left(Withthehelpofcalculator\right)$$

Commented by Tawa11 last updated on 18/Oct/21

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

Commented by MathSh last updated on 18/Oct/21

$$\mathrm{dear}\mathrm{Ser}\mathrm{ans}:1$$

Commented by mr W last updated on 18/Oct/21

$$suchquestionshavenogenerality!$$$$you{can}^{\prime }treallysolvethem.they$$$$justrequestthatyou‘‘{see}^{″}thesolution$$$$ornot.butwhenasinglevalueis$$$$changed,nosolutioncanbe‘‘{seen}^{″}.$$$$3+\frac{x}{\sqrt[3]{x^2-x+1}}=2\sqrt[3]{4(1+x)}$$$$3+\frac{1}{\sqrt[3]{\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}}}=2\sqrt[3]{4(1+x)}$$$$oneguess3+1=2\times 2,andonegetsx=1$$$$whichfulfillstheequation.sox=1$$$$isasolutionyoucansee.whether$$$$thereisanothersolution?youcan$$$$notknowandyoucannotsee.$$$$whentheeqn.ischangedto$$$$\frac{5}{\sqrt[3]{1+x}}+\frac{x}{\sqrt[3]{1+x^3}}=2\sqrt[3]{4}$$$$nosolutioncanbe‘‘{seen}^{″}.$$

Commented by mr W last updated on 18/Oct/21

$$MathShsir:$$$$x=1isnottheonlysolution!$$

Commented by MathSh last updated on 18/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}$$

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