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Question Number 171109 by akolade last updated on 08/Jun/22
Commented by Rasheed.Sindhi last updated on 09/Jun/22
a+b=2_((1)) , 3a^2 +3b^2 =4_((ii)) , a^(2019) +b^(2019) =? (i)⇒(a/b)+1=(2/b)⇒ determinant ((((a/b)=((2−b)/b)....(A)))) (ii)⇒((3a^2 )/(3b^2 ))+1=(4/(3b^2 ))⇒ determinant ((((a^2 /b^2 )=((4−3b^2 )/(3b^2 ))...(B)))) (A) & (B): ((4−3b^2 )/(3b^2 ))=(((2−b)/b))^2 ⇒3b^2 −6b+4=0 b=((6±(√(36−48)))/6)=((6±2i(√3))/6)=((3±i(√3))/3) •(A): (a/b)=((2−b)/b)=((2−((3±i(√3))/3))/((3±i(√3))/3))=((3∓i(√3))/(3±i(√3))) =(((3∓i(√3))^2 )/((3±i(√3))(3∓i(√3) )))=((9−3∓6i(√3))/(9+3)) =((1∓i(√3))/2)=−((−1±i(√3))/2)=−ω, −ω^2 (a/b)=−ω,−ω^2 ((a/b))^3 =(−ω)^3 ,(−ω^2 )^3 (a^3 /b^3 )=−1 ((a^3 /b^3 ))^(673) =(−1)^(673) (a^(2019) /b^(2019) )=−1 a^(2019) =−b^(2019) a^(2019) +b^(2019) =0
$$\underset{\left(\mathrm{1}\right)} {\underbrace{{a}+{b}=\mathrm{2}}},\:\underset{\left({ii}\right)} {\underbrace{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{4}}},\:{a}^{\mathrm{2019}} +{b}^{\mathrm{2019}} =? \\ $$$$\left({i}\right)\Rightarrow\frac{{a}}{{b}}+\mathrm{1}=\frac{\mathrm{2}}{{b}}\Rightarrow\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{\mathrm{2}−{b}}{{b}}….\left({A}\right)}\\\hline\end{array} \\ $$$$\left({ii}\right)\Rightarrow\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{3}{b}^{\mathrm{2}} }+\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}{b}^{\mathrm{2}} }\Rightarrow\begin{array}{|c|}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{4}−\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{3}{b}^{\mathrm{2}} }…\left({B}\right)}\\\hline\end{array} \\ $$$$\left({A}\right)\:\&\:\left({B}\right): \\ $$$$\frac{\mathrm{4}−\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{3}{b}^{\mathrm{2}} }=\left(\frac{\mathrm{2}−{b}}{{b}}\right)^{\mathrm{2}} \:\:\:\:\Rightarrow\mathrm{3}{b}^{\mathrm{2}} −\mathrm{6}{b}+\mathrm{4}=\mathrm{0} \\ $$$${b}=\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{48}}}{\mathrm{6}}=\frac{\mathrm{6}\pm\mathrm{2}{i}\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\bullet\left({A}\right):\:\frac{{a}}{{b}}=\frac{\mathrm{2}−{b}}{{b}}=\frac{\mathrm{2}−\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}}}{\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}}}=\frac{\mathrm{3}\mp{i}\sqrt{\mathrm{3}}}{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{3}\mp{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{3}\pm{i}\sqrt{\mathrm{3}}\right)\left(\mathrm{3}\mp{i}\sqrt{\mathrm{3}}\:\right)}=\frac{\mathrm{9}−\mathrm{3}\mp\mathrm{6}{i}\sqrt{\mathrm{3}}}{\mathrm{9}+\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}\mp{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=−\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=−\omega,\:−\omega^{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=−\omega,−\omega^{\mathrm{2}} \\ $$$$\left(\frac{{a}}{{b}}\right)^{\mathrm{3}} =\left(−\omega\right)^{\mathrm{3}} ,\left(−\omega^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }=−\mathrm{1} \\ $$$$\left(\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }\right)^{\mathrm{673}} =\left(−\mathrm{1}\right)^{\mathrm{673}} \\ $$$$\frac{{a}^{\mathrm{2019}} }{{b}^{\mathrm{2019}} }=−\mathrm{1} \\ $$$$\:{a}^{\mathrm{2019}} =−{b}^{\mathrm{2019}} \\ $$$$\:{a}^{\mathrm{2019}} +{b}^{\mathrm{2019}} =\mathrm{0} \\ $$
Answered by som(math1967) last updated on 08/Jun/22
a+b=2 ⇒(a+b)^2 =4 ⇒a^2 +b^2 +2ab=4 ⇒ (4/3) +2ab=4 [∵3a^2 +3b^2 =4 ∴a^2 +b^2 =(4/3)] 2ab= ((12−4)/3) ⇒ab=(4/3) ∴a^2 +b^2 =ab [both (4/3)] (a^2 −ab+b^2 )=0 (a+b)(a^2 −ab+b^2 )=0 a^3 +b^3 =0 a^3 =−b^3 (a^3 )^(673) =(−b^3 )^(673) a^(2019) =−b^(2019) ∴ a^(2019) +b^(2019) =0
$$\:{a}+{b}=\mathrm{2} \\ $$$$\Rightarrow\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{4} \\ $$$$\Rightarrow\:\frac{\mathrm{4}}{\mathrm{3}}\:+\mathrm{2}{ab}=\mathrm{4}\:\:\:\:\left[\because\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{4}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}}\right] \\ $$$$\mathrm{2}{ab}=\:\frac{\mathrm{12}−\mathrm{4}}{\mathrm{3}}\:\:\Rightarrow{ab}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\:\therefore{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={ab}\:\:\:\:\left[{both}\:\frac{\mathrm{4}}{\mathrm{3}}\right] \\ $$$$\:\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0} \\ $$$$\:{a}^{\mathrm{3}} =−{b}^{\mathrm{3}} \\ $$$$\:\left({a}^{\mathrm{3}} \right)^{\mathrm{673}} =\left(−{b}^{\mathrm{3}} \right)^{\mathrm{673}} \\ $$$$\:{a}^{\mathrm{2019}} =−{b}^{\mathrm{2019}} \\ $$$$\:\therefore\:\boldsymbol{{a}}^{\mathrm{2019}} +\boldsymbol{{b}}^{\mathrm{2019}} =\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Jun/22
NiCE!
$$\mathbb{N}\boldsymbol{\mathrm{i}}\mathbb{CE}! \\ $$
Commented by som(math1967) last updated on 08/Jun/22
Thanks Rasheed sir
$${Thanks}\:{Rasheed}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Jun/22
������
Commented by akolade last updated on 08/Jun/22
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jun/22
a+b=2, 3a^2 +3b^2 =4, a^(2019) +b^(2019) =? 3a^2 +3(2−a)^2 =4 3a^2 +12−12a+3a^2 −4=0 6a^2 −12a+8=0 3a^2 −6a+4=0 a=((6±(√(36−48)))/6)=((3±i(√3))/3) b=2−a=2−((3±i(√3))/3)=((3∓i(√3))/3)=a^(−) a^3 =(((3±i(√3))/3))^3 =±((8i(√3))/9) b^3 =(((3∓i(√3))/3))^3 =∓((8i(√3))/9) a^3 =−b^3 (a^3 )^(673) =(−b^3 )^(673) a^(2019) =−b^(2019) a^(2019) +b^(2019) =0
$${a}+{b}=\mathrm{2},\:\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{4},\:\:{a}^{\mathrm{2019}} +{b}^{\mathrm{2019}} =? \\ $$$$\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}−{a}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} +\mathrm{12}−\mathrm{12}{a}+\mathrm{3}{a}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{6}{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{4}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{48}}}{\mathrm{6}}=\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${b}=\mathrm{2}−{a}=\mathrm{2}−\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{3}\mp{i}\sqrt{\mathrm{3}}}{\mathrm{3}}=\overline {{a}} \\ $$$${a}^{\mathrm{3}} =\left(\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{3}} =\pm\frac{\mathrm{8}{i}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$${b}^{\mathrm{3}} =\left(\frac{\mathrm{3}\mp{i}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{3}} =\mp\frac{\mathrm{8}{i}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$${a}^{\mathrm{3}} =−{b}^{\mathrm{3}} \\ $$$$\left({a}^{\mathrm{3}} \right)^{\mathrm{673}} =\left(−{b}^{\mathrm{3}} \right)^{\mathrm{673}} \\ $$$${a}^{\mathrm{2019}} =−{b}^{\mathrm{2019}} \\ $$$${a}^{\mathrm{2019}} +{b}^{\mathrm{2019}} =\mathrm{0} \\ $$

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