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Question Number 190536 by cortano12 last updated on 05/Apr/23

 If p,q and r are the roots of equation   x^3 −3x^2 +1 = 0 then find the value   of ((3p−2))^(1/3)  +((3q−2))^(1/3) +((3r−2))^(1/3)

$$\:\mathrm{If}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\:\mathrm{of}\:\sqrt[{\mathrm{3}}]{\mathrm{3p}−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3q}−\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{3r}−\mathrm{2}}\: \\ $$

Answered by Frix last updated on 05/Apr/23

t=((3x−2))^(1/3)  ⇔ x=((t^3 +2)/3)  Inserting, transforming  t^9 −3t^6 −24t^3 −1=0  Searching for factors (using software) ⇒  t^3 +0t^2 −3t−1 [the other factor has no real roots]  ⇒ answer is 0

$${t}=\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}}\:\Leftrightarrow\:{x}=\frac{{t}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Inserting},\:\mathrm{transforming} \\ $$$${t}^{\mathrm{9}} −\mathrm{3}{t}^{\mathrm{6}} −\mathrm{24}{t}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{Searching}\:\mathrm{for}\:\mathrm{factors}\:\left(\mathrm{using}\:\mathrm{software}\right)\:\Rightarrow \\ $$$${t}^{\mathrm{3}} +\mathrm{0}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\:\left[\mathrm{the}\:\mathrm{other}\:\mathrm{factor}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\right] \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$

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