Funktsiyani nolga tenglashtiring va iloji boricha chap tomonni ko'paytiring.

\[{t^{\frac{5}{3}}} - 7{t^{\frac{4}{3}}} - 8t = t\left( {{t^{\frac{2}{3}}} - 7{t^{\frac{1}{3}}} - 8} \right) = t\left( {{t^{\frac{1}{3}}} - 8} \right)\left( {{t^{\frac{1}{3}}} + 1} \right) = 0\]

Kvadrat tenglamalarga shunchalik yopishib qolmangki, kvadrat bo'lmagan tenglamani ko'rganingizda, u bilan shug'ullana olmaysiz deb qaror qilasiz. Garchi bu funktsiya kvadratik bo'lmagan bo'lsa-da, u hali ham faktorlangan, ammo bu holatda u qadar aniq emas edi. Agar faktor bo'lgan bo'lsa, buni aniq ko'rishingiz mumkin edi.

\[t\left( {{t^2} - 7t - 8} \right)\]

E'tibor bering, yagona haqiqiy farq shundaki, ko'rsatkichlar hozir kasrdir, lekin u hali ham bir xil asosiy shaklga ega va shuning uchun faktorlarga ajratilishi mumkin.

OK, muammoga qaytish. Faktorli shakldan biz olamiz,

\[\begin{align*} \begin{aligned}t & = 0 \\ t^{\frac{1}{3}} - 8 & = 0 \\ t^{\frac{1}{3}} + 1 & = 0 \end{aligned} & \begin{aligned} \\ \hspace{0.5in} \Rightarrow \hspace{0.25in} \\ \hspace{0.25in} \Rightarrow \hspace{0.25in} \end{aligned} & \begin{aligned} \\ t^{\frac{1}{3}} & = 8 \\ t^{\frac{1}{3}} & = - 1 \end{aligned} & \begin{aligned} \\ \hspace{0.5in} \Rightarrow \hspace{0.25in} \\ \hspace{0.25in} \Rightarrow \hspace{0.25in} \end{aligned} & \begin{aligned} \\ t & = {8^3} = 512 \\ t & = {\left( { - 1} \right)^3} = - 1 \end{aligned} \end{align*}\]

Demak, funksiyaning uchta ildizi bor,

\[t = - 1,\,\,\,\,\,t = 0,\,\,\,\,\,t = 512\]