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Question Number 7076 by Tawakalitu. last updated on 09/Aug/16

Commented by Yozzii last updated on 09/Aug/16

(2+(√5))^(1/3) ={(√5)(1+(2/(√5)))}^(1/3) =5^(1/6) (1+u)^(1/3) u=(2/(√5))<1. Also (2−(√5))^(1/3) =−5^(1/6) (1−u)^(1/3) . By application of Maclaurin′s theorem, (1+u)^(1/3) =1+(1/3)u+(((1/3)(1/3−1))/(2!))u^2 +(((1/3)(1/3−1)(1/3−2))/(3!))u^3 +(((1/3)((1/3)−1)((1/3)−2)((1/3)−3))/(4!))u^4 +... (1−u)^(1/3) =1−(1/3)u+(((1/3)((1/3)−1))/(2!))u^2 −(((1/3)((1/3)−1)((1/3)−2))/(3!))u^3 +(((1/3)((1/3)−1)((1/3)−2)((1/3)−3))/(4!))u^4 +... k=5^(1/6) [(1+u)^(1/3) −(1−u)^(1/3) ]=5^(1/6) [(2/3)u+(2/(3!))((1/3))((2/3))((5/3))u^3 +(2/(5!))((1/3))((2/3))((5/3))((8/3))(((11)/3))u^5 +...] k=(2/3)5^(1/6) {(1/(1!3^0 ))u+((1×2×5)/(3!3^2 ))u^3 +((1×2×5×8×11)/(5!3^4 ))u^5 +...} k=(2/3)5^(1/6) Σ_(r=0) ^∞ (((3r+1)!!!)/(3^r (2r+1)!))u^(2r+1) (continue)

$$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\left\{\sqrt{\mathrm{5}}\left(\mathrm{1}+\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\right)\right\}^{\mathrm{1}/\mathrm{3}} =\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{1}+{u}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${u}=\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}<\mathrm{1}. \\ $$$${Also}\:\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =−\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{1}−{u}\right)^{\mathrm{1}/\mathrm{3}} . \\ $$$${By}\:{application}\:{of}\:{Maclaurin}'{s}\:{theorem}, \\ $$$$\left(\mathrm{1}+{u}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}{u}+\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(\mathrm{1}/\mathrm{3}−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} +\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(\mathrm{1}/\mathrm{3}−\mathrm{1}\right)\left(\mathrm{1}/\mathrm{3}−\mathrm{2}\right)}{\mathrm{3}!}{u}^{\mathrm{3}} +\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{3}\right)}{\mathrm{4}!}{u}^{\mathrm{4}} +... \\ $$$$\left(\mathrm{1}−{u}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}{u}+\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} −\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}\right)}{\mathrm{3}!}{u}^{\mathrm{3}} +\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{3}\right)}{\mathrm{4}!}{u}^{\mathrm{4}} +... \\ $$$${k}=\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left[\left(\mathrm{1}+{u}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{1}−{u}\right)^{\mathrm{1}/\mathrm{3}} \right]=\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left[\frac{\mathrm{2}}{\mathrm{3}}{u}+\frac{\mathrm{2}}{\mathrm{3}!}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{\mathrm{3}}\right){u}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{5}!}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\frac{\mathrm{8}}{\mathrm{3}}\right)\left(\frac{\mathrm{11}}{\mathrm{3}}\right){u}^{\mathrm{5}} +...\right] \\ $$$${k}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left\{\frac{\mathrm{1}}{\mathrm{1}!\mathrm{3}^{\mathrm{0}} }{u}+\frac{\mathrm{1}×\mathrm{2}×\mathrm{5}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }{u}^{\mathrm{3}} +\frac{\mathrm{1}×\mathrm{2}×\mathrm{5}×\mathrm{8}×\mathrm{11}}{\mathrm{5}!\mathrm{3}^{\mathrm{4}} }{u}^{\mathrm{5}} +...\right\} \\ $$$${k}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{5}^{\mathrm{1}/\mathrm{6}} \underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{3}{r}+\mathrm{1}\right)!!!}{\mathrm{3}^{{r}} \left(\mathrm{2}{r}+\mathrm{1}\right)!}{u}^{\mathrm{2}{r}+\mathrm{1}} \\ $$$$\left({continue}\right) \\ $$

Commented by Tawakalitu. last updated on 09/Aug/16

Thanks so much sir

$${Thanks}\:{so}\:{much}\:{sir} \\ $$

Answered by prakash jain last updated on 10/Aug/16

brute force method put a=1 and show equality (2+(√5))^(1/3) =x (2−(√5))^(1/3) =((−1)/x) x−(1/x)=a to prove a=1 ((x^2 −1)/x)=((x^6 −1+3x^2 −3x^4 )/x^3 )=a^3 x^6 −1+3x^2 −3x^4 =a^3 x^3 ((a^3 x^3 +1−x^6 )/(3x^2 ))=(1−x^2 ) (A) (((a^3 x^3 +1−x^6 )^3 )/(3^3 x^6 ))=1−x^6 −3x^2 +3x^4 =1−x^6 −3x^2 (1−x^2 ) (((a^3 x^3 +1−x^6 )^3 )/(3^3 x^6 ))=1−x^6 −(a^3 x^3 +1−x^6 ) Now we have all x in power of 3. so we put x=(2+(√5))^(1/3) and put a=1 and see if LHS=RHS a^3 x^3 +1−x^6 = 3+(√5)−(2+(√5))^2 =3+(√5)−4−5−4(√5)=−6−3(√5) =−3(2+(√5)) LHS= (((a^3 x^3 +1−x^6 )^3 )/(3x^6 ))= ((−3^3 (2+(√5))^3 )/(3^3 (2+(√5))^2 ))=−(2+(√5)) RHS= 1−x^6 −(a^3 x^3 +1−x^6 ) =1−(2+(√5))^2 +6+3(√5) =1−(4+5+4(√5))+6+3(√5)=−2−(√5) LHS=RHS

$$\mathrm{brute}\:\mathrm{force}\:\mathrm{method}\:\mathrm{put}\:{a}=\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{show}\:\mathrm{equality} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} ={x} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\frac{−\mathrm{1}}{{x}} \\ $$$${x}−\frac{\mathrm{1}}{{x}}={a}\:{to}\:{prove}\:{a}=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{6}} −\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{4}} }{{x}^{\mathrm{3}} }={a}^{\mathrm{3}} \\ $$$${x}^{\mathrm{6}} −\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{4}} ={a}^{\mathrm{3}} {x}^{\mathrm{3}} \\ $$$$\frac{{a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} }{\mathrm{3}{x}^{\mathrm{2}} }=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\:\:\:\left({A}\right) \\ $$$$\frac{\left({a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} {x}^{\mathrm{6}} }=\mathrm{1}−{x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{4}} =\mathrm{1}−{x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\frac{\left({a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} {x}^{\mathrm{6}} }=\mathrm{1}−{x}^{\mathrm{6}} −\left({a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} \right) \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{all}\:{x}\:\mathrm{in}\:\mathrm{power}\:\mathrm{of}\:\mathrm{3}. \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{put}\:{x}=\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} \:\mathrm{and}\:\mathrm{put}\:{a}=\mathrm{1}\:\mathrm{and}\:\mathrm{see} \\ $$$$\mathrm{if}\:\mathrm{LHS}=\mathrm{RHS} \\ $$$${a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} = \\ $$$$\mathrm{3}+\sqrt{\mathrm{5}}−\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{3}+\sqrt{\mathrm{5}}−\mathrm{4}−\mathrm{5}−\mathrm{4}\sqrt{\mathrm{5}}=−\mathrm{6}−\mathrm{3}\sqrt{\mathrm{5}} \\ $$$$=−\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$$$\mathrm{LHS}= \\ $$$$\frac{\left({a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{6}} }= \\ $$$$\frac{−\mathrm{3}^{\mathrm{3}} \left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} \left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }=−\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$$$\mathrm{RHS}= \\ $$$$\mathrm{1}−{x}^{\mathrm{6}} −\left({a}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{6}} \right) \\ $$$$=\mathrm{1}−\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{6}+\mathrm{3}\sqrt{\mathrm{5}} \\ $$$$=\mathrm{1}−\left(\mathrm{4}+\mathrm{5}+\mathrm{4}\sqrt{\mathrm{5}}\right)+\mathrm{6}+\mathrm{3}\sqrt{\mathrm{5}}=−\mathrm{2}−\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$\mathrm{LHS}=\mathrm{RHS} \\ $$

Commented by Tawakalitu. last updated on 09/Aug/16

Wow .. thanks sir

$${Wow}\:..\:{thanks}\:{sir} \\ $$

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