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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

$$\underset{\left(1\right)}{\underbrace{a+b=2}},\underset{\left(ii\right)}{\underbrace{3a^2+3b^2=4}},a^{2019}+b^{2019}=?$$$$\left(i\right)\Rightarrow \frac{a}{b}+1=\frac{2}{b}\Rightarrow \overline{)\begin{array}{c}\frac{a}{b}=\frac{2-b}{b}....\left(A\right)\end{array}}$$$$\left(ii\right)\Rightarrow \frac{3a^2}{3b^2}+1=\frac{4}{3b^2}\Rightarrow \overline{)\begin{array}{c}\frac{a^2}{b^2}=\frac{4-3b^2}{3b^2}...\left(B\right)\end{array}}$$$$\left(A\right)\mathrm{\&}\left(B\right):$$$$\frac{4-3b^2}{3b^2}={\left(\frac{2-b}{b}\right)}^2\Rightarrow 3b^2-6b+4=0$$$$b=\frac{6\pm \sqrt{36-48}}{6}=\frac{6\pm 2i\sqrt{3}}{6}=\frac{3\pm i\sqrt{3}}{3}$$$$\bullet \left(A\right):\frac{a}{b}=\frac{2-b}{b}=\frac{2-\frac{3\pm i\sqrt{3}}{3}}{\frac{3\pm i\sqrt{3}}{3}}=\frac{3\mp i\sqrt{3}}{3\pm i\sqrt{3}}$$$$=\frac{{(3\mp i\sqrt{3})}^2}{(3\pm i\sqrt{3})(3\mp i\sqrt{3})}=\frac{9-3\mp 6i\sqrt{3}}{9+3}$$$$=\frac{1\mp i\sqrt{3}}{2}=-\frac{-1\pm i\sqrt{3}}{2}=-\omega ,-{\omega }^2$$$$\frac{a}{b}=-\omega ,-{\omega }^2$$$${\left(\frac{a}{b}\right)}^3={(-\omega )}^3,{(-{\omega }^2)}^3$$$$\frac{a^3}{b^3}=-1$$$${\left(\frac{a^3}{b^3}\right)}^{673}={(-1)}^{673}$$$$\frac{a^{2019}}{b^{2019}}=-1$$$$a^{2019}=-b^{2019}$$$$a^{2019}+b^{2019}=0$$

Answered by som(math1967) last updated on 08/Jun/22

$$a+b=2$$$$\Rightarrow {(a+b)}^2=4$$$$\Rightarrow a^2+b^2+2ab=4$$$$\Rightarrow \frac{4}{3}+2ab=4[\because 3a^2+3b^2=4$$$$\therefore a^2+b^2=\frac{4}{3}]$$$$2ab=\frac{12-4}{3}\Rightarrow ab=\frac{4}{3}$$$$\therefore a^2+b^2=ab\left[both\frac{4}{3}\right]$$$$(a^2-ab+b^2)=0$$$$(a+b)(a^2-ab+b^2)=0$$$$a^3+b^3=0$$$$a^3=-b^3$$$${\left(a^3\right)}^{673}={(-b^3)}^{673}$$$$a^{2019}=-b^{2019}$$$$\therefore {\mathit{a}}^{2019}+{\mathit{b}}^{2019}=0$$

Commented by Rasheed.Sindhi last updated on 08/Jun/22

$$\mathbb{N}\mathbf{i}\mathbb{CE}!$$

Commented by som(math1967) last updated on 08/Jun/22

$$ThanksRasheedsir$$

Commented by Rasheed.Sindhi last updated on 08/Jun/22

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Commented by akolade last updated on 08/Jun/22

$$\mathrm{great}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 09/Jun/22

$$a+b=2,3a^2+3b^2=4,a^{2019}+b^{2019}=?$$$$3a^2+3{(2-a)}^2=4$$$$3a^2+12-12a+3a^2-4=0$$$$6a^2-12a+8=0$$$$3a^2-6a+4=0$$$$a=\frac{6\pm \sqrt{36-48}}{6}=\frac{3\pm i\sqrt{3}}{3}$$$$b=2-a=2-\frac{3\pm i\sqrt{3}}{3}=\frac{3\mp i\sqrt{3}}{3}=\bar{a}$$$$a^3={\left(\frac{3\pm i\sqrt{3}}{3}\right)}^3=\pm \frac{8i\sqrt{3}}{9}$$$$b^3={\left(\frac{3\mp i\sqrt{3}}{3}\right)}^3=\mp \frac{8i\sqrt{3}}{9}$$$$a^3=-b^3$$$${\left(a^3\right)}^{673}={(-b^3)}^{673}$$$$a^{2019}=-b^{2019}$$$$a^{2019}+b^{2019}=0$$

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