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Question Number 139867 by Maclaurin Stickker last updated on 01/May/21

a^3 +b^3 +c^3 =0  a^(12) +b^(12) +c^(12) =8  a^6 +b^6 +c^6 =?  (ans: 4)

$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{0} \\ $$$${a}^{\mathrm{12}} +{b}^{\mathrm{12}} +{c}^{\mathrm{12}} =\mathrm{8} \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} =? \\ $$$$\left({ans}:\:\mathrm{4}\right) \\ $$$$ \\ $$

Answered by MJS_new last updated on 02/May/21

x+y+z=0  x^4 +y^4 +z^4 =8  x^2 +y^2 +z^2 =?  z=−(x+y)  x^4 +y^4 +(x+y)^4 =8  x=u−v∧y=u+v  (u−v)^4 +(u+v)^4 +(2u)^4 =8  9u^4 +6u^2 v^2 +v^4 =4    x^2 +y^2 +(x+y)^2 =  =(u−v)^2 +(u+v)^2 +(2u)^2 =6u^2 +2v^2     (6u^2 +2v^2 )^2 =4(3u^2 +v^2 )^2 =2(9u^4 +6u^2 v^2 +v^4 )=  =4×4=16 ⇒ 6u^2 +2v^2 =4 ⇒ ... x^2 +y^2 +z^2 =4 ⇒  ⇒ a^6 +b^6 +c^6 =4

$${x}+{y}+{z}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{8} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =? \\ $$$${z}=−\left({x}+{y}\right) \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\left({x}+{y}\right)^{\mathrm{4}} =\mathrm{8} \\ $$$${x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\left({u}−{v}\right)^{\mathrm{4}} +\left({u}+{v}\right)^{\mathrm{4}} +\left(\mathrm{2}{u}\right)^{\mathrm{4}} =\mathrm{8} \\ $$$$\mathrm{9}{u}^{\mathrm{4}} +\mathrm{6}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +{v}^{\mathrm{4}} =\mathrm{4} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} = \\ $$$$=\left({u}−{v}\right)^{\mathrm{2}} +\left({u}+{v}\right)^{\mathrm{2}} +\left(\mathrm{2}{u}\right)^{\mathrm{2}} =\mathrm{6}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{6}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{3}{u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{9}{u}^{\mathrm{4}} +\mathrm{6}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +{v}^{\mathrm{4}} \right)= \\ $$$$=\mathrm{4}×\mathrm{4}=\mathrm{16}\:\Rightarrow\:\mathrm{6}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow\:...\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow \\ $$$$\Rightarrow\:{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{4} \\ $$

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