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 - \frac{1}{b} = 2 c o s \frac{2 π}{13}

b - \frac{1}{c} = 2 c o s \frac{6 π}{13}

c - \frac{1}{a} = 2 c o s \frac{8 π}{13} p r o v e t h a t

{(a b c)}^{5} - \frac{1}{{(a b c)}^{5}} = 11

Commented by mr W last updated on 30/Oct/22

l e t k = a b c

w e g e t

k - \frac{1}{k} = 1^{*) s e e l a t e r}

{(k - \frac{1}{k})}^{2} = k^{2} + \frac{1}{k^{2}} - 2 = 1

\Rightarrow k^{2} + \frac{1}{k^{2}} = 3

{(k - \frac{1}{k})}^{3} = k^{3} - \frac{1}{k^{3}} - 3 (k - \frac{1}{k}) = k^{3} - \frac{1}{k^{3}} - 3 = 1

\Rightarrow k^{3} - \frac{1}{k^{3}} = 4

(k^{3} - \frac{1}{k^{3}}) (k^{2} + \frac{1}{k^{2}}) = 4 \times 3

k^{5} - \frac{1}{k^{5}} + k - \frac{1}{k} = 4 \times 3

\Rightarrow k^{5} - \frac{1}{k^{5}} = 4 \times 3 - 1 = 11 ✓

Commented by mathlove last updated on 30/Oct/22

t h a n k s m r