Question Number 64542 by LPM last updated on 19/Jul/19
Answered by MJS last updated on 19/Jul/19
$${b}=\mathrm{1}−{a} \\ $$$${a}^{\mathrm{2}} +\left(\mathrm{1}−{a}\right)^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\pm\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}}\:\:\:{b}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\mp\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{8}} \\ $$
Commented by LPM last updated on 19/Jul/19
$$\mathrm{good} \\ $$
Answered by mr W last updated on 19/Jul/19
![a+b=1 a^2 +b^2 +2ab=1 2+2ab=1 ⇒ab=−(1/2) a^7 +b^7 =(a+b)(a^6 −a^5 b+a^4 b^2 −a^3 b^3 +a^2 b^4 −ab^5 +b^6 ) =a^6 +(1/2)a^4 +(1/4)a^2 +(1/8)+(1/4)b^2 +(1/2)b^4 +b^6 =(a^6 +b^6 )+(1/2)(a^4 +b^4 )+(1/4)(a^2 +b^2 )+(1/8) =(a^3 +b^3 )^2 −2a^3 b^3 +(1/2)[(a^2 +b^2 )^2 −2a^2 b^2 ]+(1/2)+(1/8) =(a^3 +b^3 )^2 +(1/4)+(1/2)[4−(1/2)]+(1/2)+(1/8) =(a^3 +b^3 )^2 +((21)/8) ={(a+b)[(a+b)^2 −3ab]}^2 +((21)/8) ={1+(3/2)}^2 +((21)/8) =((25)/4)+((21)/8) =((71)/8)](Q64571.png)
$${a}+{b}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{1} \\ $$$$\mathrm{2}+\mathrm{2}{ab}=\mathrm{1} \\ $$$$\Rightarrow{ab}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} =\left({a}+{b}\right)\left({a}^{\mathrm{6}} −{a}^{\mathrm{5}} {b}+{a}^{\mathrm{4}} {b}^{\mathrm{2}} −{a}^{\mathrm{3}} {b}^{\mathrm{3}} +{a}^{\mathrm{2}} {b}^{\mathrm{4}} −{ab}^{\mathrm{5}} +{b}^{\mathrm{6}} \right) \\ $$$$={a}^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{4}} +{b}^{\mathrm{6}} \\ $$$$=\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}\right]+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} +\frac{\mathrm{21}}{\mathrm{8}} \\ $$$$=\left\{\left({a}+{b}\right)\left[\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{ab}\right]\right\}^{\mathrm{2}} +\frac{\mathrm{21}}{\mathrm{8}} \\ $$$$=\left\{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right\}^{\mathrm{2}} +\frac{\mathrm{21}}{\mathrm{8}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{4}}+\frac{\mathrm{21}}{\mathrm{8}} \\ $$$$=\frac{\mathrm{71}}{\mathrm{8}} \\ $$
Commented by LPM last updated on 19/Jul/19
$$\mathrm{good} \\ $$$$ \\ $$