Question-194846

Question Number 194846 by dimentri last updated on 17/Jul/23

$$\:\:\:\:\:\:\underbrace{\:} \\ $$

Answered by som(math1967) last updated on 17/Jul/23

x=((((10+6(√3)))^(1/3) ((√3)−1))/( (√(5+1+2(√5)))−(√5))) x=((((3(√3)+3.3+3(√3)+1))^(1/3) ((√3)−1))/( (√(((√5)+1)^2 ))−(√5))) x=((((((√3))^3 +3.((√3))^2 .1+3.(√3).1+1^3 ))^(1/3) ((√3)−1))/( (√5)+1−(√5))) x=((((((√3)+1)^3 ))^(1/3) ((√3)−1))/1) x=((√3)+1)((√3)−1)=2 ∴G=(12×2^3 +4×2^2 −55)^(2023) =(96+16−55)^(2023) =(57)^(2023)

$$\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}+\mathrm{6}\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{3}.\mathrm{3}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}} +\mathrm{3}.\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} .\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{3}}.\mathrm{1}+\mathrm{1}^{\mathrm{3}} }\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}}+\mathrm{1}−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{3}} }\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{1}} \\ $$$${x}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{2} \\ $$$$\therefore{G}=\left(\mathrm{12}×\mathrm{2}^{\mathrm{3}} +\mathrm{4}×\mathrm{2}^{\mathrm{2}} −\mathrm{55}\right)^{\mathrm{2023}} \\ $$$$\:=\left(\mathrm{96}+\mathrm{16}−\mathrm{55}\right)^{\mathrm{2023}} \\ $$$$=\left(\mathrm{57}\right)^{\mathrm{2023}} \\ $$

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