Whakaoti i te (x-3)^3=125

$$( x - 3 ) ^ { 3 } = 125$$

Whakaoti mō x (complex solution)

$$x=8$$

$$x=\frac{1+5\sqrt{3}i}{2}\approx 0.5+4.330127019i$$

$$x=\frac{-5\sqrt{3}i+1}{2}\approx 0.5-4.330127019i$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} hei whakaroha \left(x-3\right)^{3}.

$$x^{3}-9x^{2}+27x-27=125$$

Tangohia te 125 mai i ngā taha e rua.

$$x^{3}-9x^{2}+27x-27-125=0$$

Tangohia te 125 i te -27, ka -152.

$$x^{3}-9x^{2}+27x-152=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -152, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±152,±76,±38,±19,±8,±4,±2,±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=8$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-9x^{2}+27x-152 ki te x-8, kia riro ko x^{2}-x+19. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}-x+19=0$$

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 -1 mō te b, me te 19 mō te c i te ture pūrua.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 19}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{-75}}{2}$$

Whakaotia te whārite x^{2}-x+19=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{-5i\sqrt{3}+1}{2} x=\frac{1+5i\sqrt{3}}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=8 x=\frac{-5i\sqrt{3}+1}{2} x=\frac{1+5i\sqrt{3}}{2}$$

Whakaoti mō x

$$x=8$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} hei whakaroha \left(x-3\right)^{3}.

$$x^{3}-9x^{2}+27x-27=125$$

Tangohia te 125 mai i ngā taha e rua.

$$x^{3}-9x^{2}+27x-27-125=0$$

Tangohia te 125 i te -27, ka -152.

$$x^{3}-9x^{2}+27x-152=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -152, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±152,±76,±38,±19,±8,±4,±2,±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=8$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-9x^{2}+27x-152 ki te x-8, kia riro ko x^{2}-x+19. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}-x+19=0$$

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 -1 mō te b, me te 19 mō te c i te ture pūrua.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 19}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{-75}}{2}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā.

$$x\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$x=8$$