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Question Number 154303 by mathdanisur last updated on 16/Sep/21

Solve in R  ((z^9  - 81z - 62)/z^3 ) = 18 ((3z + 2))^(1/3)

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{z}^{\mathrm{9}} \:-\:\mathrm{81z}\:-\:\mathrm{62}}{\mathrm{z}^{\mathrm{3}} }\:=\:\mathrm{18}\:\sqrt[{\mathrm{3}}]{\mathrm{3z}\:+\:\mathrm{2}} \\ $$

Commented by MJS_new last updated on 16/Sep/21

check the values of  y_1 =((z^9 −81z−62)/(18z^3 )); y_1 ′=((z^9 +27z+31)/(3z^4 ))  and  y_2 =((3z+2))^(1/3) ; y_2 ′=(1/((3z+2)^(2/3) ))  to see there′s no other possible solution than  x=−1∨x=2

$$\mathrm{check}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of} \\ $$$${y}_{\mathrm{1}} =\frac{{z}^{\mathrm{9}} −\mathrm{81}{z}−\mathrm{62}}{\mathrm{18}{z}^{\mathrm{3}} };\:{y}_{\mathrm{1}} '=\frac{{z}^{\mathrm{9}} +\mathrm{27}{z}+\mathrm{31}}{\mathrm{3}{z}^{\mathrm{4}} } \\ $$$$\mathrm{and} \\ $$$${y}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{3}{z}+\mathrm{2}};\:{y}_{\mathrm{2}} '=\frac{\mathrm{1}}{\left(\mathrm{3}{z}+\mathrm{2}\right)^{\mathrm{2}/\mathrm{3}} } \\ $$$$\mathrm{to}\:\mathrm{see}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{other}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{than} \\ $$$${x}=−\mathrm{1}\vee{x}=\mathrm{2} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 16/Sep/21

Try-and-Error z=2 is a solution

$$\mathrm{Try}-\mathrm{and}-\mathrm{Error}\:{z}=\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$

Commented by mathdanisur last updated on 16/Sep/21

thanks ser, yes more and z = -1

$$\mathrm{thanks}\:\mathrm{ser},\:\mathrm{yes}\:\mathrm{more}\:\mathrm{and}\:\mathrm{z}\:=\:-\mathrm{1} \\ $$

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