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Question Number 67371 by TawaTawa last updated on 26/Aug/19

Answered by mind is power last updated on 26/Aug/19

a^2 3b=b^3 +9(√3)...1 b^2 3a=a^3 −10....2 3bc^2 =b^3 +45(√3)...3 3cb^2 =c^3 +28...4 3ac^2 =a^3 −88...5 3ca^2 =c^3 +16...6 ..1−i..2⇒b^3 −ia^3 +9(√3)+10i=3ba^2 −3iab^2 ⇒b^3 −ia^3 −3ba^2 +3(ia)b=−9(√3)−10i ⇒(b+ia)^3 =−9(√3)−10i=7(√7)(((−9(√3))/(7(√7)))−((10i)/(7(√7))))=7(√7)e^(i(((b+2kπ)/))) sam idea giv us (a+ic)^3 =88−16i=8(11−2i)=40(√5)(((11)/(5(√5)))−(2/(5(√5)))i)=(2(√5))^3 e^(ia+2kπ) ⇒ a+ic=(2(√5))e^(i(((a+2kπ)/3))) bee continued

$${a}^{\mathrm{2}} \mathrm{3}{b}={b}^{\mathrm{3}} +\mathrm{9}\sqrt{\mathrm{3}}...\mathrm{1} \\ $$$${b}^{\mathrm{2}} \mathrm{3}{a}={a}^{\mathrm{3}} −\mathrm{10}....\mathrm{2} \\ $$$$\mathrm{3}{bc}^{\mathrm{2}} ={b}^{\mathrm{3}} +\mathrm{45}\sqrt{\mathrm{3}}...\mathrm{3} \\ $$$$\mathrm{3}{cb}^{\mathrm{2}} ={c}^{\mathrm{3}} +\mathrm{28}...\mathrm{4} \\ $$$$\mathrm{3}{ac}^{\mathrm{2}} ={a}^{\mathrm{3}} −\mathrm{88}...\mathrm{5} \\ $$$$\mathrm{3}{ca}^{\mathrm{2}} ={c}^{\mathrm{3}} +\mathrm{16}...\mathrm{6} \\ $$$$..\mathrm{1}−{i}..\mathrm{2}\Rightarrow{b}^{\mathrm{3}} −{ia}^{\mathrm{3}} +\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{10}{i}=\mathrm{3}{ba}^{\mathrm{2}} −\mathrm{3}{iab}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{3}} −{ia}^{\mathrm{3}} −\mathrm{3}{ba}^{\mathrm{2}} +\mathrm{3}\left({ia}\right){b}=−\mathrm{9}\sqrt{\mathrm{3}}−\mathrm{10}{i} \\ $$$$\Rightarrow\left({b}+{ia}\right)^{\mathrm{3}} =−\mathrm{9}\sqrt{\mathrm{3}}−\mathrm{10}{i}=\mathrm{7}\sqrt{\mathrm{7}}\left(\frac{−\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{7}\sqrt{\mathrm{7}}}−\frac{\mathrm{10}{i}}{\mathrm{7}\sqrt{\mathrm{7}}}\right)=\mathrm{7}\sqrt{\mathrm{7}}{e}^{{i}\left(\frac{{b}+\mathrm{2}{k}\pi}{}\right)} \\ $$$${sam}\:{idea}\:{giv}\:{us}\: \\ $$$$ \\ $$$$\left({a}+{ic}\right)^{\mathrm{3}} =\mathrm{88}−\mathrm{16}{i}=\mathrm{8}\left(\mathrm{11}−\mathrm{2}{i}\right)=\mathrm{40}\sqrt{\mathrm{5}}\left(\frac{\mathrm{11}}{\mathrm{5}\sqrt{\mathrm{5}}}−\frac{\mathrm{2}}{\mathrm{5}\sqrt{\mathrm{5}}}{i}\right)=\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{3}} {e}^{{ia}+\mathrm{2}{k}\pi} \\ $$$$\Rightarrow\:{a}+{ic}=\left(\mathrm{2}\sqrt{\mathrm{5}}\right){e}^{{i}\left(\frac{{a}+\mathrm{2}{k}\pi}{\mathrm{3}}\right)} \\ $$$${bee}\:{continued}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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