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Question Number 177530 by mnjuly1970 last updated on 06/Oct/22

$$$$$$f\left(x\right)+f\left(\frac{1}{\sqrt[3]{1-x^3}}\right)=x^3$$$$isgiven.findthevalueof:$$$$$$$$f(-1)=?$$$$$$

Answered by floor(10²Eta[1]) last updated on 06/Oct/22

$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\sqrt[3]{1-{\mathrm{x}}^3}}\right)={\mathrm{x}}^3\left(1\right)$$$$\mathrm{x}\to \frac{1}{\sqrt[3]{1-{\mathrm{x}}^3}}$$$$\mathrm{f}\left(\frac{1}{\sqrt[3]{1-{\mathrm{x}}^3}}\right)+\mathrm{f}\left(\frac{\sqrt[3]{{\mathrm{x}}^3-1}}{\mathrm{x}}\right)=\frac{1}{1-{\mathrm{x}}^3}\left(2\right)$$$$\mathrm{x}\to \frac{1}{\sqrt[3]{1-{\mathrm{x}}^3}}$$$$\mathrm{f}\left(\frac{\sqrt[3]{{\mathrm{x}}^3-1}}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^3}\left(3\right)$$$$\left(3\right)-\left(2\right):$$$$\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\frac{1}{\sqrt[3]{1-{\mathrm{x}}^3}}\right)=\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^3}-\frac{1}{1-{\mathrm{x}}^3}\left(4\right)$$$$\left(1\right)+\left(4\right):$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}(\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^3}-\frac{1}{1-{\mathrm{x}}^3}+{\mathrm{x}}^3)$$$$\Rightarrow \mathrm{f}(-1)=\frac{1}{4}$$$$$$

Commented by mnjuly1970 last updated on 07/Oct/22

$$thankssir$$

Answered by mr W last updated on 06/Oct/22

$$f(-1)+f\left(\frac{1}{\sqrt[3]{2}}\right)=-1...\left(i\right)$$$$f\left(\frac{1}{\sqrt[3]{2}}\right)+f\left(\sqrt[3]{2}\right)=\frac{1}{2}...\left(ii\right)$$$$f\left(\sqrt[3]{2}\right)+f(-1)=2...\left(iii\right)$$$$\left(i\right)+\left(iii\right)-\left(ii\right):$$$$2f(-1)=2-1-\frac{1}{2}=\frac{1}{2}$$$$\Rightarrow f(-1)=\frac{1}{4}$$

Commented by mnjuly1970 last updated on 07/Oct/22

$$thanksalotsirW$$

Answered by a.lgnaoui last updated on 06/Oct/22

$$pourx{=}^3\sqrt{2}$$$$f\left({}^3\sqrt{2}\right)+f\frac{1}{{}^3\sqrt{1-2}}=2\left(dapreslexpressiin\right)$$$$doncf\left({}^3\sqrt{2}\right)+f(-1)=2\left(1\right)$$$$pourx=-1$$$$f(-1)+f\left(\frac{1}{{}^3\sqrt{1-{(-1)}^3}}\right)=f(-1)+f\left(\frac{1}{{}^3\sqrt{2}}\right)=-1\left(2\right)$$$$\left(1\right)et\left(2\right)\Rightarrow 2-f\left({}^3\sqrt{2}\right)=-1-f\left(\frac{1}{{}^3\sqrt{2}}\right)$$$$soitf\left({}^3\sqrt{2}\right)-f\left(\frac{1}{{}^3\sqrt{2}}\right)=3\left(3\right)$$$$Calculdef\left(\frac{1}{{}^3\sqrt{2}}\right)?$$$$f\left(\frac{1}{{}^3\sqrt{2}}\right)+f\left(\frac{1}{{}^3\sqrt{1-\frac{1}{2}}}\right)=\frac{1}{2}$$$$f\left(\frac{1}{{}^3\sqrt{2}}\right)+f\left({}^3\sqrt{2}\right)=\frac{1}{2}\left(4\right)$$$$\left(3\right)et\left(4\right)\Rightarrow f\left(\frac{1}{{}^3\sqrt{2}}\right)=f\left({}^3\sqrt{2}\right)-3=\frac{1}{2}-f(-1)$$$$\left(3\right)et\left(4\right)\Rightarrow \{\begin{array}{l}f\left({}^3\sqrt{2}\right)-f\left(\frac{1}{{}^3\sqrt{2}}\right)=3\\ f\left({}^3\sqrt{2}\right)+f\left(\frac{1}{{}^3\sqrt{2}}\right)=\frac{1}{2}\end{array}$$$$2f\left({}^3\sqrt{2}\right)=\frac{7}{2}f\left(\sqrt{2}\right)=\frac{7}{4}$$$$finalementen[remplacantdans\left(1\right)$$$$f\left(\sqrt{2}\right)+f(-1)=2$$$$f(-1)=2-f\left({}^3\sqrt{2}\right)=2-\frac{7}{4}$$$$f(-1)=\frac{1}{4}$$

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