Whakaoti i te x^3+1/x^3=3

$${ x }^{ 3 } + \frac{ 1 }{ { x }^{ 3 } } =3$$

Whakaoti mō x (complex solution)

$$x\in \frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}e^{\frac{2\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}e^{\frac{2\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x^{3}.

$$x^{3}x^{3}+1=3x^{3}$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 3 me te 3 kia riro ai te 6.

$$x^{6}+1=3x^{3}$$

Tangohia te 3x^{3} mai i ngā taha e rua.

$$x^{6}+1-3x^{3}=0$$

Whakakapia te t mō te x^{3}.

$$t^{2}-3t+1=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -3 mō te b, me te 1 mō te c i te ture pūrua.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{3±\sqrt{5}}{2}$$

Whakaotia te whārite t=\frac{3±\sqrt{5}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

Mai i te x=t^{3}, ka taea ngā otinga mā te whakaoti te whārite mō ia t.

$$x=-\sqrt[3]{\frac{\sqrt{5}+3}{2}}e^{\frac{\pi i}{3}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}}ie^{\frac{\pi i}{6}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=-\sqrt[3]{\frac{3-\sqrt{5}}{2}}e^{\frac{\pi i}{3}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}ie^{\frac{\pi i}{6}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=\sqrt[3]{\frac{3-\sqrt{5}}{2}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}ie^{\frac{\pi i}{6}}\text{, }x\neq 0 x=-\sqrt[3]{\frac{3-\sqrt{5}}{2}}e^{\frac{\pi i}{3}}\text{, }x\neq 0 x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}}ie^{\frac{\pi i}{6}}\text{, }x\neq 0 x=-\sqrt[3]{\frac{\sqrt{5}+3}{2}}e^{\frac{\pi i}{3}}\text{, }x\neq 0$$

Whakaoti mō x

$$x=\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}}{2}\approx 0.72556263$$

$$x = \frac{2 ^ {\frac{2}{3}} \sqrt[3]{\sqrt{5} + 3}}{2} \approx 1.378240772$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x^{3}.

$$x^{3}x^{3}+1=3x^{3}$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 3 me te 3 kia riro ai te 6.

$$x^{6}+1=3x^{3}$$

Tangohia te 3x^{3} mai i ngā taha e rua.

$$x^{6}+1-3x^{3}=0$$

Whakakapia te t mō te x^{3}.

$$t^{2}-3t+1=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -3 mō te b, me te 1 mō te c i te ture pūrua.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{3±\sqrt{5}}{2}$$

Whakaotia te whārite t=\frac{3±\sqrt{5}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

I te mea ko x=t^{3}, ka riro ngā otinga mā te arotake i te x=\sqrt[3]{t} mō ia t.

$$x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}$$