a(b+(1/b))=−1 ....(i) b−(1/b)=a^2 ....(ii) Solve simultaneously for a, and b.


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Question Number 36221 by ajfour last updated on 30/May/18

$a (b + \frac{1}{b}) = - 1 . . . . (i)$ $b - \frac{1}{b} = a^{2} . . . . (i i)$ $S o l v e s i m u l t a n e o u s l y f o r a, a n d b .$
Commented by behi83417@gmail.com last updated on 30/May/18

$a^{2} (b^{2} + b^{- 2} + 2) = 1 \Rightarrow (b - b^{- 1}) (b^{2} + b^{- 2} + 2) = 1$ $b^{3} + b^{- 1} + 2 b - b - b^{- 3} - 2 b^{- 1} = 1$ $b^{3} + b - b^{- 1} - b^{- 3} = 1 \Rightarrow b^{6} + b^{4} - b^{2} - 1 = b^{3}$ $\Rightarrow b^{6} + b^{4} - b^{3} - b^{2} - 1 = 0 . . . .$
	Answered by Rasheed.Sindhi last updated on 30/May/18

	$a (b + \frac{1}{b}) = - 1 . . . . (i)$ $b - \frac{1}{b} = a^{2} . . . . (i i)$ $(i) \Rightarrow b + \frac{1}{b} = - 1 / a$ ${(b + \frac{1}{b})}^{2} = {(- 1 / a)}^{2}$ $\Rightarrow b^{2} + \frac{1}{b^{2}} = \frac{1}{a^{2}} - 2 . . . . . . . . . . (iii)$ $(ii) \Rightarrow {(b - \frac{1}{b})}^{2} = {(a^{2})}^{2}$ $\Rightarrow b^{2} + \frac{1}{b^{2}} = a^{4} + 2 . . . . . . . . . . . . (iv)$ $(iii) & (iv) : a^{4} + 2 = \frac{1}{a^{2}} - 2$ $a^{6} + 4 a^{2} - 1 = 0$ $(i) \Rightarrow a {(b + \frac{1}{b})}^{2} = \frac{1}{a^{2}} . . . . . . . . . . . (v)$ $(ii) \times (v) :$ $(b - \frac{1}{b}) {(b + \frac{1}{b})}^{2} = 1$ $(b^{2} - \frac{1}{b^{2}}) (b + \frac{1}{b}) = 1$ $b^{3} - \frac{1}{b^{3}} + b - \frac{1}{b} = 1$ ${(b - \frac{1}{b})}^{3} + 4 (b - \frac{1}{b}) = 1$ $C o n t i n u e$
	Commented by MJS last updated on 30/May/18

	$a^{6} + 4 a^{2} - 1 = 0$ $a = \sqrt{t}$ $t^{3} + 4 t - 1 = 0$ $solution almost the same as of the former$ $problem (w^{3} + 4 w + 1 = 0)$ $t = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} =$ $= \sqrt[3]{\frac{1}{2} + \frac{\sqrt{849}}{18}} + \sqrt[3]{\frac{1}{2} - \frac{\sqrt{849}}{18}}$
	Commented by ajfour last updated on 30/May/18

	$t h a n k y o u s i r!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 30/May/18

	$a^{2} (b^{2} + 2 + \frac{1}{b^{2}}) = 1$ $(b - \frac{1}{b}) (b^{2} + 2 + \frac{1}{b^{2}}) = 1$ $(\frac{b^{2} - 1}{b}) (\frac{b^{4} + 2 b^{2} + 1}{b^{2}}) = 1$ $b^{6} + 2 b^{4} + b^{2} - b^{4} - 2 b^{2} - 1 = b^{3}$ $b^{6} + b^{4} - b^{3} - b^{2} - 1 = 0$ $b^{3} + b - 1 - \frac{1}{b} - \frac{1}{b^{3}} = 0$ $(b^{3} - \frac{1}{b^{3}}) + (b - \frac{1}{b}) - 1 = 0$ $b - \frac{1}{b} = k$ ${(b - \frac{1}{b})}^{3} = k^{3}$ $b^{3} - 3 b^{2} . \frac{1}{b} + 3 b . \frac{1}{b^{2}} - \frac{1}{b^{3}} = k^{3}$ $b^{3} - \frac{1}{b^{3}} - 3 (b - \frac{1}{b}) = k^{3}$ $s o b^{3} - \frac{1}{b^{3}} = k^{3} + 3 k$ $s o e a u a t i o n i s$ $k^{3} + 3 k + k - 1 = 0$ $k^{3} + 4 k - 1 = 0$ $k = \frac{z}{n}$ ${(\frac{z}{n})}^{3} + 4 (\frac{z}{n}) - 1 = 0$ $z^{3} + 4 {z n}^{2} - n^{3} = 0$ $c o s 3 θ = 4 {c o s}^{3} θ - 3 c o s θ$ ${c o s}^{3} θ - \frac{1}{4} c o s 3 θ - \frac{3}{4} c o s θ = 0$
	Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18

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