3xy-2x-y-3bx-3c-0-3xy-x-2y-3by-3c-0-find-non-zero-real-values-of-x-y-if-b-c-R-

Question Number 79613 by ajfour last updated on 26/Jan/20

3 x y (2 x - y) - 3 b x + 3 c = 0

3 x y (x - 2 y) - 3 b y - 3 c = 0

f i n d n o n - z e r o, r e a l v a l u e s

o f x, y i f b, c \in R .

Answered by behi83417@gmail.com last updated on 26/Jan/20

xy (2 x - y) = bx - c

xy (x - 2 y) = by + c

\Rightarrow {\begin{cases} b (x + y) = 3 xy (x - y) \\ xy (x + y) + 2 c = b (x - y) \end{cases}

\Rightarrow {\begin{cases} 1 . x - y = 0 \Rightarrow x = y \Rightarrow x^{3} - bx - c = 0 \\ 2 . \frac{b (x + y)}{xy (x + y) + 2 c} = \frac{3 xy}{b} \end{cases}

[x^{3} - bx - c = 0, △ = B^{2} - 3 AC = 3 b

k = \frac{27 c}{2 \sqrt{{27 b}^{3}}} = \frac{9 c}{2 b \sqrt{3 b}}

1 . if : b > 0, ∣ k ∣⩽ 1 \Rightarrow {\begin{cases} x_{1} = 2 \sqrt{\frac{b}{3}} \cos (\frac{\cos^{- 1} (\frac{9 c}{2 b \sqrt{3 b}})}{3}) \\ x_{2, 3} = 2 \sqrt{\frac{b}{3}} \cos (\frac{\cos^{- 1} (\frac{9 c}{2 b \sqrt{3 b}}) \pm \frac{2 π}{3}}{3}) \end{cases}

2 . if : b > 0, k > 1 \Rightarrow x = y = \sqrt{\frac{b}{3}} . [\sqrt[3]{\frac{9 ∣ c ∣}{2 b \sqrt{3 b}} + \sqrt{\frac{{27 c}^{2}}{{4 b}^{3}} - 1}} + \sqrt[3]{\frac{9 ∣ c ∣}{2 b \sqrt{3 b}} - \sqrt{\frac{{27 c}^{2}}{{4 b}^{3}} - 1}}]

3 . if : b < 0 \Rightarrow x = y = \sqrt{\frac{∣ b ∣}{3}} . [\sqrt[3]{\frac{9 c}{2 b \sqrt{3 ∣ b ∣}} + \sqrt{\frac{{27 c}^{2}}{4 ∣ b^{3} ∣} + 1}} + \sqrt[3]{\frac{9 c}{2 b \sqrt{3 ∣ b ∣}} - \sqrt{\frac{{27 c}^{2}}{4 ∣ b^{3} ∣} + 1}}]

4 . b = 0 \Rightarrow x = y = \sqrt[3]{c}

- - - - - - - - - - + - - - - - - -

\frac{b (x + y)}{xy (x + y) + 2 c} = \frac{3 xy}{b} \Rightarrow

b^{2} (x + y) = 3 {(xy)}^{2} (x + y) + 6 cxy

\Rightarrow 3 (x + y) {(xy)}^{2} + 6 cxy - b (x + y) = 0

\Rightarrow (x + y) [b - {3 x}^{2} y^{2}] = 6 cxy \overset{xy = p}{\Rightarrow} x + y = \frac{6 cp}{b - {3 p}^{2}}

\Rightarrow b^{2} {(\frac{6 cp}{b - {3 p}^{2}})}^{2} = {9 p}^{2} [{(\frac{6 cp}{b - {3 p}^{2}})}^{2} - 4 p]

\Rightarrow {36 c}^{2} p^{2} b^{2} = {9 p}^{2} [{36 c}^{2} p^{2} - 4 p {(b - {3 p}^{2})}^{2}]

[\Rightarrow p = 0 \Rightarrow no answer for x and y]

\Rightarrow c^{2} b^{2} = {9 c}^{2} p^{2} - {pb}^{2} + {6 bp}^{2} - p^{5}

\Rightarrow p^{5} - 3 (2 b + {3 c}^{2}) p^{2} + b^{2} p + c^{2} b^{2} = 0

i {can}^{'} t solve this \dots \dots

Commented by ajfour last updated on 26/Jan/20

T h a n k s S i r, f o r s u c h a g e n e r a l

s o l u t i o n . .

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