Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 10381 by ridwan balatif last updated on 06/Feb/17

Commented by ridwan balatif last updated on 06/Feb/17

is there any simple solution to solve this question?

$$\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question}? \\ $$

Answered by mrW1 last updated on 06/Feb/17

let t=x^(1/6)   ⇒x=t^6   x^(1/3) =t^2   x^(1/2) =t^3   x^(2/3) =t^4   x^(4/3) =t^8     (((x^(1/3) −x^(1/6) )(x^(1/2) +x)(x^(1/2) +x^(1/3) +x^(2/3) ))/((x^(4/3) −x)(x+x^(1/3) +x^(2/3) )))  =(((t^2 −t)(t^3 +t^6 )(t^3 +t^2 +t^4 ))/((t^8 −t^6 )(t^6 +t^2 +t^4 )))  =(((t−1)(1+t^3 )(t+1+t^2 )t^6 )/((t^2 −1)(t^4 +1+t^2 )t^8 ))  =(((t^3 +1)(t^2 +t+1))/((t+1)(t^4 +t^2 +1)t^2 ))  =(((t+1)(t^2 −t+1)(t^2 +t+1))/((t+1)(t^4 +t^2 +1)t^2 ))  =((t^4 +t^2 +1)/((t^4 +t^2 +1)t^2 ))  =(1/t^2 )=(1/x^(1/3) )=x^(−(1/3))     ⇒ answer (A)

$${let}\:{t}={x}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\Rightarrow{x}={t}^{\mathrm{6}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{3}}} ={t}^{\mathrm{2}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} ={t}^{\mathrm{3}} \\ $$$${x}^{\frac{\mathrm{2}}{\mathrm{3}}} ={t}^{\mathrm{4}} \\ $$$${x}^{\frac{\mathrm{4}}{\mathrm{3}}} ={t}^{\mathrm{8}} \\ $$$$ \\ $$$$\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)\left({x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{x}\right)\left({x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}{\left({x}^{\frac{\mathrm{4}}{\mathrm{3}}} −{x}\right)\left({x}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)} \\ $$$$=\frac{\left({t}^{\mathrm{2}} −{t}\right)\left({t}^{\mathrm{3}} +{t}^{\mathrm{6}} \right)\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right)}{\left({t}^{\mathrm{8}} −{t}^{\mathrm{6}} \right)\left({t}^{\mathrm{6}} +{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right)} \\ $$$$=\frac{\left({t}−\mathrm{1}\right)\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\left({t}+\mathrm{1}+{t}^{\mathrm{2}} \right){t}^{\mathrm{6}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}+{t}^{\mathrm{2}} \right){t}^{\mathrm{8}} } \\ $$$$=\frac{\left({t}^{\mathrm{3}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }={x}^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$ \\ $$$$\Rightarrow\:{answer}\:\left({A}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com