Question

Este $\sqrt[3]{5+\sqrt{10}}+\sqrt[3]{5-\sqrt{10}}$ soluție a ecuației x⁹ - 30x⁶ - 105x³ - 1000 = 0?

class: X.

Answer 1

$\displaystyle Notam~t= \sqrt[3]{5+ \sqrt{10}}+ \sqrt[3]{5- \sqrt{10}}. \\ \\ In~baza~identitatii~(a+b)^3=a^3+b^3+3ab(a+b),~pentru \\ \\ a=\sqrt[3]{5+ \sqrt{10}}~si~b=\sqrt[3]{5- \sqrt{10}},~obtinem: \\ \\ t^3=5+ \sqrt{10}+5- \sqrt{10}+3 \cdot \sqrt[3]{(5+ \sqrt{10})(5- \sqrt{10})} \cdot t \Leftrightarrow \\ \\ \Leftrightarrow t^3=10+3 \sqrt[3]{25-10} \cdot t \Leftrightarrow$

$\displaystyle \boxed{t^3=10+3 \sqrt[3]{15} \cdot t} ~~~(*) \\ \\ Ridicam~relatia~(*)~la~patrat \Rightarrow \boxed{t^6=100+60 \sqrt[3]{15} \cdot t + 9 \sqrt[3]{225} \cdot t^2}. \\ \\ Ridicam~(*)~la~cub \Rightarrow \\ \\ t^9=1000+405t^3+90 \sqrt[3]{15}t(10+3 \sqrt[3]{15}t). \\ \\ Inlocuim~t^3~cu~10+3 \sqrt[3]{15} \cdot t.~Dupa~efectuarea~calculelor~obtinem \\ \\ \boxed{t^9=5050+2115 \sqrt[3]{15}t+270 \sqrt[3]{225} t^2}.$

$\displaystyle Inlocuim~t^3,~t^6~si~t^9~cu~expresiile~din~chenare~in~expresia \\ \\ x^9-30x^6-105^3-1000. \\ \\ Toti~termenii~se~vor~reduce,~rezultatul~fiind~0. \\ \\ Deci~t~este~solutie~a~ecuatiei~date.$