Question Number 212181 by RojaTaniya last updated on 05/Oct/24
Answered by A5T last updated on 05/Oct/24
![a^2 +b^2 +c^2 =(a+b+c)^2 −2(ab+bc+ca)=5 ⇒ab+bc+ca=2 a^3 +b^3 +c^3 −3abc=(a+b+c)[(a+b+c)^2 −3(ab+bc+ca)] ⇒27−3abc=3[9−3(2)]=9⇒abc=6 ⇒a,b,c are roots of x^3 −3x^2 +2x−6=0 ⇒x^2 (x−3)+2(x−3)=(x−3)(x^2 +2)=0 (a,b,c)=(3,i(√2),−i(√2)) upto permutation ⇒a^(100) +b^(100) +c^(100) =3^(100) +2^(50) +2^(50) =3^(100) +2^(51)](https://www.tinkutara.com/question/Q212184.png)
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{5} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{2} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\left({a}+{b}+{c}\right)\left[\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{3}\left({ab}+{bc}+{ca}\right)\right] \\ $$$$\Rightarrow\mathrm{27}−\mathrm{3}{abc}=\mathrm{3}\left[\mathrm{9}−\mathrm{3}\left(\mathrm{2}\right)\right]=\mathrm{9}\Rightarrow{abc}=\mathrm{6} \\ $$$$\Rightarrow{a},{b},{c}\:{are}\:{roots}\:{of}\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({x}−\mathrm{3}\right)+\mathrm{2}\left({x}−\mathrm{3}\right)=\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{3},{i}\sqrt{\mathrm{2}},−{i}\sqrt{\mathrm{2}}\right)\:{upto}\:{permutation} \\ $$$$\Rightarrow{a}^{\mathrm{100}} +{b}^{\mathrm{100}} +{c}^{\mathrm{100}} =\mathrm{3}^{\mathrm{100}} +\mathrm{2}^{\mathrm{50}} +\mathrm{2}^{\mathrm{50}} =\mathrm{3}^{\mathrm{100}} +\mathrm{2}^{\mathrm{51}} \\ $$
Commented by RojaTaniya last updated on 05/Oct/24
$${Sir}\:{thanks} \\ $$