p-3-q-3-c-pq-1-3-0-lt-c-lt-2-3-3-Find-p-q-

Question Number 187388 by ajfour last updated on 16/Feb/23

p^{3} + q^{3} = c, p q = \frac{1}{3} \forall 0 < c < \frac{2}{3 \sqrt{3}}

F i n d p + q .

Answered by anurup last updated on 17/Feb/23

{(p + q)}^{3} - 3 p q (p + q) = c

x^{3} - x - c = 0, x = p + q

x = u + v

\Rightarrow x^{3} = u^{3} + v^{3} + 3 u v x

\Rightarrow x^{3} - 3 u v x - (u^{3} + v^{3}) = 0

∴ u v = \frac{1}{3} and u^{3} + v^{3} = c

{(u^{3} - v^{3})}^{2} = {(u^{3} + v^{3})}^{2} - 4 u^{3} v^{3}

\Rightarrow (u^{3} - v^{3}) = \pm \sqrt{c^{2} - \frac{4}{27}}

∴ u^{3} = \frac{1}{2} (c \pm \sqrt{c^{2} - \frac{4}{27}}), v^{3} = \frac{1}{2} (c \mp \sqrt{c^{2} - \frac{4}{27}})

∵ u^{3} v^{3} = \frac{1}{27}, u^{3} = \frac{1}{2} (c + \sqrt{c^{2} - \frac{4}{27}}), v^{3} = \frac{1}{2} (c - \sqrt{c^{2} - \frac{4}{27}})

∵ 0 < c < \frac{2}{3 \sqrt{3}}, c^{2} < \frac{4}{27}

∴ u^{3} = \frac{1}{2} (c + i \sqrt{\frac{4}{27} - c^{2}}), v^{3} = \frac{1}{2} (c - i \sqrt{\frac{4}{27} - c^{2}})

u = \frac{1}{\sqrt{3}} {(\cos θ + i \sin θ)}^{\frac{1}{3}}, v = \frac{1}{\sqrt{} 3} {(\cos θ - i \sin θ)}^{\frac{1}{3}} where \tan θ = \frac{\sqrt{\frac{4}{27} - c^{2}}}{c}

u = \frac{1}{\sqrt{} 3} {\cos \frac{(θ + 2 k π)}{3} + i \sin \frac{(θ + 2 k π)}{3}}, v = \frac{1}{\sqrt{3}} {\cos \frac{(θ + 2 k π)}{3} - i \sin \frac{(θ + 2 k π)}{3}}, k = 0, 1, 2

For each value of k we will get an ordered pair (u, v)

each of which corresponds to a value of x,

θ = \tan^{- 1} \frac{\sqrt{\frac{4}{27} - c^{2}}}{c}

Commented by anurup last updated on 17/Feb/23

waiting for your feedback

Commented by ajfour last updated on 17/Feb/23

t h a n k s, s e e i h a v e t a k e n a n o t h e r

w a y \dots

Answered by ajfour last updated on 17/Feb/23

B a s i c a l l y x^{3} = x + c

I f x = p + q

p^{3} + q^{3} + 3 p q (p + q) = p + q + c

I f w e t a k e p^{3} + q^{3} = c \Rightarrow

p q = \frac{1}{3} a s p + q = x \neq 0

l e t a n o t h e r w a y x = \frac{\sqrt{3}}{2} (\frac{\cos ϕ}{\cos θ})

\Rightarrow 3 \sqrt{3} \cos^{3} ϕ = (\sqrt{3} \cos ϕ) {(2 \cos θ)}^{2}

+ 8 c \cos^{3} θ

f r o m 4 \cos^{3} ϕ - 3 \cos ϕ = \cos 3 ϕ

w e c o m p a r e c o e f f i c i e n t s

\frac{3 \sqrt{3}}{4 \sqrt{3} \cos^{2} θ} = \frac{4}{3}

\Rightarrow \cos^{2} θ = \frac{9}{16}

f i r s t s i m p l y l e t \cos θ = \frac{3}{4}

\Rightarrow \frac{3 \sqrt{3}}{4} \cos 3 ϕ = 8 c \cos^{3} θ

\cos 3 ϕ = \frac{4}{3 \sqrt{3}} (8 c) {(\frac{3}{4})}^{3} = \frac{3 \sqrt{3} c}{2}

\Rightarrow 3 ϕ = 2 k π \pm \cos^{- 1} (\frac{3 \sqrt{3} c}{2})

\Rightarrow x = p + q = \frac{\sqrt{3}}{2} \times \frac{4}{3} \cos ϕ

x = \frac{2}{\sqrt{3}} \cos [\frac{2 k π}{3} \pm \frac{1}{3} \cos^{- 1} (\frac{3 \sqrt{3} c}{2})]

k = 0, 1, 2

s a y o n e p + q = x i s t h e n f o r k = 0

x = p + q = \frac{2}{\sqrt{3}} \cos (\frac{1}{3} \cos^{- 1} (\frac{3 \sqrt{3} c}{2}))

Commented by anurup last updated on 17/Feb/23

wow! Innovative, but I just want to know what

makes you think to assume x = \frac{\sqrt{3}}{2} (\frac{\cos ϕ}{\cos θ})

Commented by ajfour last updated on 17/Feb/23

i h a v e t r i e d 3500 s u b s t i t u t i o n s!

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