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Question Number 208652 by efronzo1 last updated on 20/Jun/24

Answered by Berbere last updated on 20/Jun/24

$$a,bsolutionof-3x^3+2x=c$$$$S_1={\int }_0^{a}c-(2x-3x^3)={\int }_a^b(2x-3x^3-c)dx$$$$=ca-a^2-\frac{3}{4}{a}^4=b^2-a^2-\frac{3}{4}(b^4-a^4)-c(b-a)$$$$b(b-\frac{3}{4}{b}^3-c)=0$$$$c=b-\frac{3}{4}{b}^3$$$$\Rightarrow 2b-3b^3=b-\frac{3}{4}{b}^3$$$$\Rightarrow b-\frac{9}{4}{b}^3=0\Rightarrow b^2=\frac{4}{9}\Rightarrow b=\frac{2}{3}$$$$c=\frac{2}{3}-\frac{3}{4}{\left(\frac{2}{3}\right)}^3=\frac{2}{3}-\frac{2}{9}=\frac{4}{9}$$$$-3x^3+2x-\frac{4}{9}$$$$x^3-\frac{2}{3}x+\frac{4}{27}=$$$$(x-\frac{2}{3})(x^2+\frac{2}{3}x-\frac{2}{9})=0$$$$\frac{12}{9};\frac{-\frac{2}{3}+\sqrt{\frac{12}{9}}}{2}=\frac{\sqrt{3}-1}{3}=a$$$$a=\frac{\sqrt{3}-1}{3};b=\frac{2}{3};c=\frac{4}{9}$$$$$$

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