Question Number 151317 by mathdanisur last updated on 20/Aug/21
$$\mathrm{if}\:\:\boldsymbol{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:-\:\sqrt{\mathrm{6}}}\:+\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:+\:\sqrt{\mathrm{6}}} \\ $$$$\mathrm{find}\:\:\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{4}} \:+\:\mathrm{9x}^{\mathrm{2}} \:=\:? \\ $$
Answered by iloveisrael last updated on 20/Aug/21
$$\:{we}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}}\:=\:\mathrm{0} \\ $$$${implies}\:{that}\:{x}^{\mathrm{3}} −\left(\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}\right)−\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}\right)=\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{6}}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{7}}\:=\:\mathrm{3}{x}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{x}\:=\:\mathrm{2}\sqrt{\mathrm{7}}\: \\ $$$$\Rightarrow\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} \:=\:\mathrm{28} \\ $$$$\Rightarrow{x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{2}} =\:\mathrm{28}\: \\ $$
Answered by som(math1967) last updated on 20/Aug/21
![let a=^3 (√((√7)−(√6))) ,b=^3 (√((√7)+(√6))) ∴ab=^3 (√(((√7)−(√6))((√7)+(√6)))) =^3 (√(7−6))=1 a^3 +b^3 =(√7)−(√6)+(√7)+(√6)=2(√7) now x=a+b x^3 =(a+b)^3 x^3 =a^3 +b^3 +3ab(a+b) x^3 =2(√7)+3.1.x [x=a+b] x^3 −3x=2(√7) (x^3 −3x)^2 =(2(√7))^2 x^6 −6x^4 +9x^2 =28 ans](https://www.tinkutara.com/question/Q151318.png)
$$\boldsymbol{{let}}\:\boldsymbol{{a}}=^{\mathrm{3}} \sqrt{\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}}\:,\boldsymbol{{b}}=^{\mathrm{3}} \sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}} \\ $$$$\therefore\boldsymbol{{ab}}=^{\mathrm{3}} \sqrt{\left(\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}\right)\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=^{\mathrm{3}} \sqrt{\mathrm{7}−\mathrm{6}}=\mathrm{1} \\ $$$$\:\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} =\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}+\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}=\mathrm{2}\sqrt{\mathrm{7}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{x}}=\boldsymbol{{a}}+\boldsymbol{{b}} \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} =\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{3}} \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} =\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} +\mathrm{3}\boldsymbol{{ab}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right) \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} =\mathrm{2}\sqrt{\mathrm{7}}+\mathrm{3}.\mathrm{1}.\boldsymbol{{x}}\:\:\:\left[\boldsymbol{{x}}=\boldsymbol{{a}}+\boldsymbol{{b}}\right] \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{x}}=\mathrm{2}\sqrt{\mathrm{7}} \\ $$$$\left(\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{x}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{x}}^{\mathrm{6}} −\mathrm{6}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{9}\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{28}\:\boldsymbol{{ans}} \\ $$
Commented by mathdanisur last updated on 20/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$