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Question Number 153226 by ajfour last updated on 05/Sep/21

Commented by mr W last updated on 05/Sep/21

$$x^3+x-c=0$$$$\Rightarrow x=\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}+\frac{c}{2}}-\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}-\frac{c}{2}}$$$$anyotherwaystosolve?$$

Commented by ajfour last updated on 05/Sep/21

Commented by TVTA last updated on 06/Sep/21

$$x^3+x-c=0$$$$\Rightarrow x=\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}+\frac{c}{2}}-\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}-\frac{c}{2}}$$$$anyotherwaystosolve?$$$$x=\sqrt[3]{\frac{c+\sqrt{c^2+\frac{4}{27}}}{2}}+\frac{1}{\sqrt[3]{\frac{c+\sqrt{c^2+\frac{4}{27}}}{2}}}$$$$isthiscorrectandoksir?$$

Commented by ajfour last updated on 06/Sep/21

$$thankseveryone!$$

Answered by talminator2856791 last updated on 06/Sep/21

$$2x(x^2-1)+2x=c$$$$2x^3-2x+2x=c$$$$2x^3=c$$$$x^3=c\cdot 2^{-1}$$$$x={(c\cdot 2^{-1})}^{\frac{1}{3}}$$

Commented by MJS_new last updated on 06/Sep/21

$$\mathrm{formula}\mathrm{for}\mathrm{area}\mathrm{of}\mathrm{trapezoid}\mathrm{with}\mathrm{sides}$$$$a\parallel c\mathrm{and}d=b\mathrm{with}\mathrm{given}\mathrm{height}h\mathrm{is}\frac{(a+c)h}{2}$$$$\mathrm{in}\mathrm{this}\mathrm{case}$$$$\frac{(2+2x^2)x}{2}=c\iff x^3+x=c$$

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