z^6-z^3+1=0 eching

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

z uchun yechish

$$z\in e^{\frac{5\pi i}{9}},e^{\frac{17\pi i}{9}},e^{\frac{11\pi i}{9}},e^{\frac{13\pi i}{9}},e^{\frac{\pi i}{9}},e^{\frac{7\pi i}{9}}$$

Kvadrat tenglamalaridan foydalanish qoidalari

z^{3} uchun t ni almashtiring.

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

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun -1 ni va c uchun 1 ni ayiring.

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

Hisoblarni amalga oshiring.

$$t=\frac{1±\sqrt{-3}}{2}$$

t=\frac{1±\sqrt{-3}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

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

z=t^{3} boʻlganda, yechimlar har bir t uchun tenglamani yechish orqali olinadi.

$$z=-e^{\frac{4\pi i}{9}} z=ie^{\frac{5\pi i}{18}} z=e^{\frac{\pi i}{9}} z=-ie^{\frac{7\pi i}{18}} z=-e^{\frac{2\pi i}{9}} z=ie^{\frac{\pi i}{18}}$$