p^3 +q^3 =c , pq=(1/3) ∀ 0<c<(2/(3(√3))) Find p+q.


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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
	$(p+q)^3 −3pq(p+q)=c x^3 −x−c=0,x=p+q x=u+v ⇒x^3 =u^3 +v^3 +3uvx ⇒x^3 −3uvx−(u^3 +v^3 )=0 ∴uv=(1/3) and u^3 +v^3 =c (u^3 −v^3 )^2 =(u^3 +v^3 )^2 −4u^3 v^3 ⇒(u^3 −v^3 )=±(√(c^2 −(4/(27)))) ∴u^3 =(1/2)(c±(√(c^2 −(4/(27)))) ),v^3 =(1/2)(c∓(√(c^2 −(4/(27))))) ∵u^3 v^3 =(1/(27)), u^3 =(1/2)(c+(√(c^2 −(4/(27)))) ),v^3 =(1/2)(c−(√(c^2 −(4/(27)))) ) ∵0<c<(2/(3(√3))), c^2 <(4/(27)) ∴u^3 =(1/2)(c+i(√((4/(27))−c^2 )) ),v^3 =(1/2)(c−i(√((4/(27))−c^2 )) ) u=(1/( (√3)))(cos θ+isin θ)^(1/3) , v=(1/( (√)3))(cos θ−isin θ)^(1/3) where tan θ=((√((4/(27))−c^2 ))/c) u=(1/( (√)3)){cos (((θ+2kπ))/3)+isin (((θ+2kπ))/3)},v=(1/( (√3))){cos (((θ+2kπ))/3)−isin (((θ+2kπ))/3)},k=0,1,2 For each value of k we will get an ordered pair(u,v) each of which corresponds to a value of x, θ=tan^(−1) ((√((4/(27))−c^2 ))/c)$
	${(p + q)}^{3} - 3 p q (p + q) = c$ $x^{3} - x - c = 0, x = p + q$ $x = u + v$ $\Rightarrow x^{3} = u^{3} + v^{3} + 3 u v x$ $\Rightarrow x^{3} - 3 u v x - (u^{3} + v^{3}) = 0$ $∴ u v = \frac{1}{3} and u^{3} + v^{3} = c$ ${(u^{3} - v^{3})}^{2} = {(u^{3} + v^{3})}^{2} - 4 u^{3} v^{3}$ $\Rightarrow (u^{3} - v^{3}) = \pm \sqrt{c^{2} - \frac{4}{27}}$ $∴ u^{3} = \frac{1}{2} (c \pm \sqrt{c^{2} - \frac{4}{27}}), v^{3} = \frac{1}{2} (c \mp \sqrt{c^{2} - \frac{4}{27}})$ $∵ u^{3} v^{3} = \frac{1}{27}, u^{3} = \frac{1}{2} (c + \sqrt{c^{2} - \frac{4}{27}}), v^{3} = \frac{1}{2} (c - \sqrt{c^{2} - \frac{4}{27}})$ $∵ 0 < c < \frac{2}{3 \sqrt{3}}, c^{2} < \frac{4}{27}$ $∴ u^{3} = \frac{1}{2} (c + i \sqrt{\frac{4}{27} - c^{2}}), v^{3} = \frac{1}{2} (c - i \sqrt{\frac{4}{27} - c^{2}})$ $u = \frac{1}{\sqrt{3}} {(\cos θ + i \sin θ)}^{\frac{1}{3}}, v = \frac{1}{\sqrt{} 3} {(\cos θ - i \sin θ)}^{\frac{1}{3}} where \tan θ = \frac{\sqrt{\frac{4}{27} - c^{2}}}{c}$ $u = \frac{1}{\sqrt{} 3} {\cos \frac{(θ + 2 k π)}{3} + i \sin \frac{(θ + 2 k π)}{3}}, v = \frac{1}{\sqrt{3}} {\cos \frac{(θ + 2 k π)}{3} - i \sin \frac{(θ + 2 k π)}{3}}, k = 0, 1, 2$ $For each value of k we will get an ordered pair (u, v)$ $each of which corresponds to a value of x,$ $θ = \tan^{- 1} \frac{\sqrt{\frac{4}{27} - c^{2}}}{c}$
	Commented byanurup last updated on 17/Feb/23

	$waiting for your feedback$
	Commented byajfour last updated on 17/Feb/23

	$t h a n k s, s e e i h a v e t a k e n a n o t h e r$ $w a y . . .$
	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 byanurup 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 byajfour 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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