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a-1-b-2cos-2pi-13-b-1-c-2cos-6pi-13-c-1-a-2cos-8pi-13-prove-that-abc-5-1-abc-5-11-




Question Number 179518 by mathlove last updated on 30/Oct/22
a−(1/b)=2cos((2π)/(13))  b−(1/c)=2cos((6π)/(13))  c−(1/a)=2cos((8π)/(13))      prove that  (abc)^5 −(1/((abc)^5 ))=11
$${a}−\frac{\mathrm{1}}{{b}}=\mathrm{2}{cos}\frac{\mathrm{2}\pi}{\mathrm{13}} \\ $$$${b}−\frac{\mathrm{1}}{{c}}=\mathrm{2}{cos}\frac{\mathrm{6}\pi}{\mathrm{13}} \\ $$$${c}−\frac{\mathrm{1}}{{a}}=\mathrm{2}{cos}\frac{\mathrm{8}\pi}{\mathrm{13}}\:\:\:\:\:\:{prove}\:{that} \\ $$$$\left({abc}\right)^{\mathrm{5}} −\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{5}} }=\mathrm{11} \\ $$
Commented by mr W last updated on 30/Oct/22
let k=abc  we get  k−(1/k)=1   ^(∗) see later)   (k−(1/k))^2 =k^2 +(1/k^2 )−2=1  ⇒k^2 +(1/k^2 )=3  (k−(1/k))^3 =k^3 −(1/k^3 )−3(k−(1/k))=k^3 −(1/k^3 )−3=1  ⇒k^3 −(1/k^3 )=4  (k^3 −(1/k^3 ))(k^2 +(1/k^2 ))=4×3  k^5 −(1/k^5 )+k−(1/k)=4×3  ⇒k^5 −(1/k^5 )=4×3−1=11 ✓
$${let}\:{k}={abc} \\ $$$${we}\:{get} \\ $$$${k}−\frac{\mathrm{1}}{{k}}=\mathrm{1}\:\:\:\:^{\left.\ast\right)\:{see}\:{later}} \\ $$$$\left({k}−\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} ={k}^{\mathrm{2}} +\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{2}=\mathrm{1} \\ $$$$\Rightarrow{k}^{\mathrm{2}} +\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\left({k}−\frac{\mathrm{1}}{{k}}\right)^{\mathrm{3}} ={k}^{\mathrm{3}} −\frac{\mathrm{1}}{{k}^{\mathrm{3}} }−\mathrm{3}\left({k}−\frac{\mathrm{1}}{{k}}\right)={k}^{\mathrm{3}} −\frac{\mathrm{1}}{{k}^{\mathrm{3}} }−\mathrm{3}=\mathrm{1} \\ $$$$\Rightarrow{k}^{\mathrm{3}} −\frac{\mathrm{1}}{{k}^{\mathrm{3}} }=\mathrm{4} \\ $$$$\left({k}^{\mathrm{3}} −\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)\left({k}^{\mathrm{2}} +\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)=\mathrm{4}×\mathrm{3} \\ $$$${k}^{\mathrm{5}} −\frac{\mathrm{1}}{{k}^{\mathrm{5}} }+{k}−\frac{\mathrm{1}}{{k}}=\mathrm{4}×\mathrm{3} \\ $$$$\Rightarrow{k}^{\mathrm{5}} −\frac{\mathrm{1}}{{k}^{\mathrm{5}} }=\mathrm{4}×\mathrm{3}−\mathrm{1}=\mathrm{11}\:\checkmark \\ $$
Commented by mathlove last updated on 30/Oct/22
thanks mr
$${thanks}\:{mr} \\ $$

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