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Question Number 126109 by mnjuly1970 last updated on 17/Dec/20

....nice calculus... verify that :: A=(((2+(√5)))^(1/3) /(1+(√5))) is a rational number ...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:{calculus}... \\ $$$$\:\:{verify}\:\:{that}\:::\:\:\mathrm{A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:{is} \\ $$$$\:\:\:\:\:\:{a}\:\:{rational}\:\:{number}\:... \\ $$

Answered by som(math1967) last updated on 17/Dec/20

(((8(2+(√(5)))))^(1/3) /((1+(√5))×2)) =(((16+8(√5)))^(1/3) /(2(1+(√5)))) =(((1^3 +3×1^2 ×(√5)+3×1×((√5))^2 +((√5))^3 ))^(1/3) /(2(1+(√5)))) =((((1+(√5))^3 ))^(1/3) /(2(1+(√5))))=(1/2) is a rational number.

$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}\left(\mathrm{2}+\sqrt{\left.\mathrm{5}\right)}\right.}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)×\mathrm{2}} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{16}+\mathrm{8}\sqrt{\mathrm{5}}}}{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}^{\mathrm{3}} +\mathrm{3}×\mathrm{1}^{\mathrm{2}} ×\sqrt{\mathrm{5}}+\mathrm{3}×\mathrm{1}×\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }}{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }}{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{is}\:{a}\:{rational} \\ $$$${number}. \\ $$

Commented by mnjuly1970 last updated on 17/Dec/20

thanks alot

$${thanks}\:{alot} \\ $$

Answered by Dwaipayan Shikari last updated on 17/Dec/20

A=(((1+t))^(1/3) /t)=(√((1/t^3 )+(1/t^2 ))) =(((1/((1+(√5))^3 ))+(1/((1+(√5))^2 ))))^(1/3) =((((((√5)−1)/4))^2 +((((√5)−1)/4))^3 ))^(1/3) =((((5(√5)−1−3(√5)((√5)−1))/(64))+((24−8(√5))/(64))))^(1/3) =(((5(√5)−1−15+3(√5)+24−8(√5))/(64)))^(1/3) =(1/2)

$${A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+{t}}}{{t}}=\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }} \\ $$$$\:=\sqrt[{\mathrm{3}}]{\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}\sqrt{\mathrm{5}}−\mathrm{1}−\mathrm{3}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{64}}+\frac{\mathrm{24}−\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{64}}} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}\sqrt{\mathrm{5}}−\mathrm{1}−\mathrm{15}+\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{24}−\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{64}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 17/Dec/20

grateful ...

$${grateful}\:... \\ $$

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