Question Number 93953 by mhmd last updated on 16/May/20
Answered by Rasheed.Sindhi last updated on 16/May/20
$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{25} \\ $$$$\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{27} \\ $$$$\:\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\mathrm{27}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{27}^{\mathrm{2}} −\mathrm{2}=\mathrm{727} \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} =\left(\mathrm{727}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\left(\mathrm{727}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{528527} \\ $$
Answered by Kunal12588 last updated on 16/May/20
$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{5} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{25}+\mathrm{2}=\mathrm{27} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{4}/\mathrm{28}/\mathrm{49}=\mathrm{729}−\mathrm{2}=\mathrm{727} \\ $$$$\Rightarrow{x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\mathrm{49}/\mathrm{28}/\mathrm{102}/\mathrm{28}/\mathrm{49}=\mathrm{528529}−\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\mathrm{528527} \\ $$
Commented by Rasheed.Sindhi last updated on 16/May/20
$$\mathcal{S}{ir}/\mathcal{M}{iss}\:\mathcal{K}{unal}! \\ $$$${What}\:{are}\:{the}\:{numbers}?: \\ $$$$\:\:\:\:\:\:\mathrm{4}/\mathrm{28}/\mathrm{49} \\ $$$$\:\:\:\:\:\:\mathrm{49}/\mathrm{28}/\mathrm{102}/\mathrm{28}/\mathrm{49} \\ $$
Commented by Kunal12588 last updated on 17/May/20
$${i}\:{was}\:{squaring}\:\mathrm{27}\:{and}\:\mathrm{727}\:{inline}\:{without} \\ $$$${calculator};\:{vedic}\:{maths} \\ $$$${for}\:{any}\:\mathrm{2}−{digit}\:{no}. \\ $$$${suppose}\:{AB} \\ $$$$\left({AB}\right)^{\mathrm{2}} ={A}^{\mathrm{2}} /\mathrm{2}×{A}×{B}/{B}^{\mathrm{2}} \\ $$$${every}\:/\:/\:{slots}\:{give}\:{only}\:{one}\:{single}\:{digit} \\ $$$${eg}.\:\mathrm{34}^{\mathrm{2}} =\mathrm{9}/\mathrm{2}×\mathrm{3}×\mathrm{4}/\mathrm{16} \\ $$$$=\mathrm{9}/\mathrm{24}/\mathrm{16} \\ $$$${since}\:{we}\:{have}\:\mathrm{2}−{digit}\:{in}\:{right}\:{slot} \\ $$$${we}\:{can}\:{carry}\:{over}\:{the}\:{extra}\:{digit} \\ $$$$=\mathrm{9}/\mathrm{24}+\mathrm{1}/\mathrm{6} \\ $$$$=\mathrm{9}/\mathrm{25}/\mathrm{6} \\ $$$$=\mathrm{9}+\mathrm{2}/\mathrm{5}/\mathrm{6} \\ $$$$=\mathrm{11}/\mathrm{5}/\mathrm{6} \\ $$$$=\mathrm{1156}\:{answer} \\ $$$${with}\:{practice}\:{we}\:{can}\:{do}\:{it}\:{in}\:{single}\:{line} \\ $$$${and}\:{much}\:{faster};\:{here}\:{they}\:{don}'{t}\:{allow} \\ $$$${calculator}\:{in}\:{school}\:{so}\:{we}\:{learn}\:{some}\:{tricks}. \\ $$$$\left({ABC}\right)^{\mathrm{2}} ={A}^{\mathrm{2}} /\mathrm{2}\centerdot{A}\centerdot{B}/{B}^{\mathrm{2}} +\mathrm{2}\centerdot{A}\centerdot{C}/\mathrm{2}\centerdot{B}\centerdot{C}/{C}^{\mathrm{2}} \\ $$$${BTW},\:{I}\:{am}\:{Male},\:{Kunal}\:{is}\:{normal}\:{Indian} \\ $$$${male}\:{name}. \\ $$
Commented by Rasheed.Sindhi last updated on 19/Jun/20
$$\mathcal{T}{hank}\:{you}\:{Kunal}\:\boldsymbol{{sir}}\:{for}\:{detailed} \\ $$$${clear}\:\&{easy}\:{to}\:{understand} \\ $$$${answer}.{I}\:{think}\:{you}\:{may}\:{be}\:{a}\:{good} \\ $$$${teacher}.{Some}\:{time}\:{ago}\:{I}\:{had}\:{got} \\ $$$${a}\:{small}\:{book}:\:\boldsymbol{{Vedic}}\:\boldsymbol{{mathematics}}. \\ $$$${I}\:{liked}\:{it}\:{very}\:{much}\:{but}\:{it}\:{is}\:{matter} \\ $$$${of}\:{sorrow}\:{that}\:{I}\:{couldn}'{t}\:{be}\:'{used}-{to}' \\ $$$${of}\:{the}\:{tricks}\:{of}\:{book}! \\ $$$${Sorry}\:{for}\:{writing}\:'{miss}'\:{for}\:{you} \\ $$