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Question Number 46414 by Meritguide1234 last updated on 25/Oct/18

Commented by math1967 last updated on 25/Oct/18

Is the answar 2(√3)

$${Is}\:{the}\:{answar}\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Answered by math1967 last updated on 25/Oct/18

let (√3) +(√2) =a ∴(√3) −(√2) =(1/a) ∴x= ^5 (√(a )) +(1/(a)^(1/5) )=p+(1/p) [let ^5 (√a) =p] x^3 =p^3 +(1/p^3 ) +3(p+(1/p)) x^3 −3x=p^3 +(1/p^3 ) [∵p+(1/p)=x] now x^5 =p^5 +(1/p^5 ) +5(p^3 +(1/p^3 ))+10(p+(1/p)) x^5 =p^5 +(1/p^5 ) +5(x^3 −3x)+10x x^5 −5x^3 +5x= (√3) +(√2) +(1/((√3) −(√2))) ∴ x^3 −5x^3 +5x=(√3) +(√2) +(√3) −(√2)=2(√3)

$${let}\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:={a}\:\therefore\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\:=\frac{\mathrm{1}}{{a}} \\ $$$$\therefore{x}=\overset{\mathrm{5}} {\:}\sqrt{{a}\:}\:+\frac{\mathrm{1}}{\sqrt[{\mathrm{5}}]{{a}}}={p}+\frac{\mathrm{1}}{{p}}\:\:\left[{let}\overset{\mathrm{5}} {\:}\sqrt{{a}}\:={p}\right] \\ $$$$\:\:{x}^{\mathrm{3}} ={p}^{\mathrm{3}} +\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:+\mathrm{3}\left({p}+\frac{\mathrm{1}}{{p}}\right) \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}={p}^{\mathrm{3}} +\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:\:\:\:\left[\because{p}+\frac{\mathrm{1}}{{p}}={x}\right] \\ $$$${now}\:{x}^{\mathrm{5}} ={p}^{\mathrm{5}} +\frac{\mathrm{1}}{{p}^{\mathrm{5}} }\:+\mathrm{5}\left({p}^{\mathrm{3}} +\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\right)+\mathrm{10}\left({p}+\frac{\mathrm{1}}{{p}}\right) \\ $$$${x}^{\mathrm{5}} ={p}^{\mathrm{5}} +\frac{\mathrm{1}}{{p}^{\mathrm{5}} }\:+\mathrm{5}\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)+\mathrm{10}{x} \\ $$$${x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{3}} +\mathrm{5}{x}=\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}} \\ $$$$\therefore\:{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{3}} +\mathrm{5}{x}=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by Meritguide1234 last updated on 26/Oct/18

nice solution

$${nice}\:{solution} \\ $$

Commented by math1967 last updated on 26/Oct/18

Thank you sir

$${Thank}\:{you}\:{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Oct/18

x=a^(1/5) +a^(−(1/5)) x^5 −5x^3 +5x (a^(1/5) +a^((−1)/5) )^5 −5(a^(1/5) +a^((−1)/5) )^3 +5(a^(1/5) +a^((−1)/5) ) =a+5c_1 a^(4/5) a^((−1)/5) +5c_2 a^(3/5) a^((−2)/5) +5c_3 a^(2/5) a^((−3)/5) +5c_4 a^(1/5) a^((−4)/5) +5c_5 a^(−1) −5(a^(3/5) +3c_1 a^(2/5) a^((−1)/5) +3c_2 a^(1/5) a^((−2)/5) +a^((−3)/(5))) +5(a^(1/5) +a^((−1)/5) ) =a+5a^(3/5) +10a^(1/5) +10a^((−1)/5) +5a^((−3)/5) +(1/a)+ −5a^(3/5) −15a^(1/5) −15a^((−1)/5) −5a^((−3)/5) +5a^(1/5) +5^((−1)/5) =a+(1/a)=(√3) +(√2) +(√3) −(√2) =2(√3) answer

$${x}={a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{a}^{−\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$${x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{3}} +\mathrm{5}{x} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} \right)^{\mathrm{5}} −\mathrm{5}\left({a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} \right)^{\mathrm{3}} +\mathrm{5}\left({a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} \right) \\ $$$$={a}+\mathrm{5}{c}_{\mathrm{1}} {a}^{\frac{\mathrm{4}}{\mathrm{5}}} {a}^{\frac{−\mathrm{1}}{\mathrm{5}}} +\mathrm{5}{c}_{\mathrm{2}} {a}^{\frac{\mathrm{3}}{\mathrm{5}}} {a}^{\frac{−\mathrm{2}}{\mathrm{5}}} +\mathrm{5}{c}_{\mathrm{3}} {a}^{\frac{\mathrm{2}}{\mathrm{5}}} {a}^{\frac{−\mathrm{3}}{\mathrm{5}}} +\mathrm{5}{c}_{\mathrm{4}} {a}^{\frac{\mathrm{1}}{\mathrm{5}}} {a}^{\frac{−\mathrm{4}}{\mathrm{5}}} +\mathrm{5}{c}_{\mathrm{5}} {a}^{−\mathrm{1}} \\ $$$$−\mathrm{5}\left({a}^{\frac{\mathrm{3}}{\mathrm{5}}} +\mathrm{3}{c}_{\mathrm{1}} {a}^{\frac{\mathrm{2}}{\mathrm{5}}} {a}^{\frac{−\mathrm{1}}{\mathrm{5}}} +\mathrm{3}{c}_{\mathrm{2}} {a}^{\frac{\mathrm{1}}{\mathrm{5}}} {a}^{\frac{−\mathrm{2}}{\mathrm{5}}} +{a}^{\frac{−\mathrm{3}}{\left.\mathrm{5}\right)}} \right. \\ $$$$+\mathrm{5}\left({a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} \right) \\ $$$$={a}+\mathrm{5}{a}^{\frac{\mathrm{3}}{\mathrm{5}}} +\mathrm{10}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{10}{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} +\mathrm{5}{a}^{\frac{−\mathrm{3}}{\mathrm{5}}} +\frac{\mathrm{1}}{{a}}+ \\ $$$$\:\:\:−\mathrm{5}{a}^{\frac{\mathrm{3}}{\mathrm{5}}} −\mathrm{15}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} −\mathrm{15}{a}^{\frac{−\mathrm{1}}{\mathrm{5}}} −\mathrm{5}{a}^{\frac{−\mathrm{3}}{\mathrm{5}}} +\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{5}^{\frac{−\mathrm{1}}{\mathrm{5}}} \\ $$$$={a}+\frac{\mathrm{1}}{{a}}=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\:=\mathrm{2}\sqrt{\mathrm{3}}\:{answer} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Meritguide1234 last updated on 26/Oct/18

very good Mr.Tanmoy Are you professor?

$${very}\:{good}\:{Mr}.{Tanmoy} \\ $$$${Are}\:{you}\:{professor}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Oct/18

no sir..i enjoy it...i have interest in physics and mathematics...

$${no}\:{sir}..{i}\:{enjoy}\:\:{it}...{i}\:{have}\:{interest}\:{in}\:{physics} \\ $$$${and}\:{mathematics}... \\ $$

Commented by Meritguide1234 last updated on 26/Oct/18

basically i am teacher of senior secondary school of taiwan.

$${basically}\:{i}\:{am}\:{teacher}\:{of}\:{senior}\:{secondary}\:{school}\:{of}\:{taiwan}. \\ $$

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