Whakaoti i te (x-1)^3=54/2

$$( x - 1 ) ^ { 3 } = \frac { 54 } { 2 }$$

Whakaoti mō x (complex solution)

$$x=4$$

$$x=\frac{-1+3\sqrt{3}i}{2}\approx -0.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i-1}{2}\approx -0.5-2.598076211i$$

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-1\right)^{3}.

$$x^{3}-3x^{2}+3x-1=\frac{54}{2}$$

Whakawehea te 54 ki te 2, kia riro ko 27.

$$x^{3}-3x^{2}+3x-1=27$$

Tangohia te 27 mai i ngā taha e rua.

$$x^{3}-3x^{2}+3x-1-27=0$$

Tangohia te 27 i te -1, ka -28.

$$x^{3}-3x^{2}+3x-28=0$$

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 -28, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±28,±14,±7,±4,±2,±1$$

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.

$$x=4$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-3x^{2}+3x-28 ki te x-4, kia riro ko x^{2}+x+7. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}+x+7=0$$

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 7 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 7}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{-27}}{2}$$

Whakaotia te whārite x^{2}+x+7=0 ina he tōrunga te ±, ina he tōraro te ±.

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

Rārangitia ngā otinga katoa i kitea.

$$x=4 x=\frac{-3i\sqrt{3}-1}{2} x=\frac{-1+3i\sqrt{3}}{2}$$

Whakaoti mō x

$$x=4$$

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-1\right)^{3}.

$$x^{3}-3x^{2}+3x-1=\frac{54}{2}$$

Whakawehea te 54 ki te 2, kia riro ko 27.

$$x^{3}-3x^{2}+3x-1=27$$

Tangohia te 27 mai i ngā taha e rua.

$$x^{3}-3x^{2}+3x-1-27=0$$

Tangohia te 27 i te -1, ka -28.

$$x^{3}-3x^{2}+3x-28=0$$

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 -28, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±28,±14,±7,±4,±2,±1$$

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.

$$x=4$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-3x^{2}+3x-28 ki te x-4, kia riro ko x^{2}+x+7. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}+x+7=0$$

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 7 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 7}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{-27}}{2}$$

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.

$$x=4$$