Determine-a-b-c-in-terms-of-ab-c-bc-a-ca-b-

Question Number 59581 by Rasheed.Sindhi last updated on 12/May/19

D e t e r m i n e a, b, c i n t e r m s o f α, β, γ .

a b + c = γ

b c + a = α

c a + b = β

Answered by mr W last updated on 13/May/19

i f α = β = γ = 0, \Rightarrow a = b = c = 0 o r - 1

i f α = β = γ = 2, \Rightarrow a = b = c = 1 o r - 2

i f α = β = γ = k, \Rightarrow a = b = c = \frac{- 1 \pm \sqrt{1 + 4 k}}{2}

o t h e r w i s e :

a^{2} b + a c = a γ

- (i i i) :

(a^{2} - 1) b = a γ - β

\Rightarrow b = \frac{a γ - β}{a^{2} - 1}

a^{2} c + a b = a β

- (i) :

(a^{2} - 1) c = a β - γ

c = \frac{a β - γ}{a^{2} - 1}

p u t i n t o (i i) :

\frac{(a β - γ) (a γ - β)}{{(a^{2} - 1)}^{2}} + a = α

β γ a^{2} - (β^{2} + γ^{2}) a + β γ + a^{5} - 2 a^{3} + a - α a^{4} + 2 α a^{2} - α = 0

\Rightarrow a^{5} - α a^{4} - 2 a^{3} + (2 α + β γ) a^{2} - (1 + β^{2} + γ^{2}) a - (α - β γ) = 0

\dots \dots ?

Answered by ajfour last updated on 12/May/19

c = γ - ab

b (γ - ab) + a = α

a (γ - ab) + b = β

a + b = \frac{α + β}{1 + γ - ab}

a - b = \frac{α - β}{1 - γ + ab}

{(a + b)}^{2} - {(a - b)}^{2} = 4 ab

\Rightarrow \frac{{(α + β)}^{2}}{{(1 + γ - ab)}^{2}} + \frac{{(α - β)}^{2}}{{(1 - γ + ab)}^{2}} = 4 ab

let ab = x, α + β = s, α - β = d

s^{2} {(1 - γ + x)}^{2} + d^{2} {(1 + γ - x)}^{2} = 4 x {(1 + γ - x)}^{2} {(1 - γ + x)}^{2}

quintic, ab = x

Now 2 a = \frac{s}{1 + γ - x} + \frac{d}{1 - γ + x}

2 b = \frac{s}{1 + γ - x} - \frac{d}{1 - γ + x} .

c = γ - x .

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