if-x-3-y-3-3xy-1-then-x-y-

Question Number 174059 by mr W last updated on 23/Jul/22

i f x^{3} + y^{3} + 3 x y = 1, t h e n x + y = ?

Commented by infinityaction last updated on 23/Jul/22

y = 0 a n d x = 1

t h e n x + y = 1

o r x = - 1 a n d y = - 1

t h e n x + y = - 2

Commented by Tawa11 last updated on 24/Jul/22

Great sirs

Commented by dragan91 last updated on 24/Jul/22

(x + y) (x^{2} - xy + y^{2}) = 1 - 3 xy

(x + y) ({(x + y)}^{2} - 3 xy) = 1 - 3 xy

{(x + y)}^{3} - 1 - 3 xy (x + y) + 3 xy = 0

(x + y - 1) ({(x + y)}^{2} + x + y + 1) - 3 xy (x + y - 1) = 0

(x + y - 1) (x^{2} + 2 xy + y^{2} + x + y + 1 - 3 xy) = 0

1) x + y = 1

2) x^{2} - xy + y^{2} + x + y + 1 = 0

x^{2} + x (1 - y) + y^{2} + y + 1 = 0

x_{1, 2} = \frac{y - 1 \pm \sqrt{1 - 2 y + y^{2} - {4 y}^{2} - 4 y - 4}}{2}

= \frac{y - 1}{2} \pm \frac{\sqrt{- 3 (y^{2} + 2 y + 1)}}{2}

= \frac{y - 1}{2} \pm \frac{(y + 1) i \sqrt{3}}{2}

\Rightarrow x + y = \frac{3 y - 1}{2} \pm \frac{(y + 1) i \sqrt{3}}{2}, y \in R

Answered by behi834171 last updated on 24/Jul/22

x + y = p, x y = q

(x + y) [{(x + y)}^{2} - 3 x y] + 3 x y = 1

\Rightarrow p (p^{2} - 3 q) + 3 q = 1 \Rightarrow p^{3} - 3 p q + 3 q - 1 = 0

(p - 1) (p^{2} + p - 3 q + 1) = 0

\Rightarrow p = 1, p = \frac{- 1 \pm \sqrt{12 q - 3}}{2}

\Rightarrow x + y = p = 1 . [o n l y p o s s i b i l i t y, i t h i n k]

[12 q - 3 = 0 \Rightarrow q = \frac{1}{4}, p = - \frac{1}{2}

\Rightarrow z^{2} + \frac{z}{2} + \frac{1}{4} = 0 \Rightarrow 4 z^{2} + 2 z + 1 = 0

z = \frac{- 2 \pm \sqrt{4 - 16}}{8} = \frac{- 2 \pm 2 \sqrt{3} i}{8} = - (\frac{1}{4} \mp i . \frac{\sqrt{3}}{4})

\Rightarrow {\begin{cases} x = - (\frac{1}{4} + i . \frac{\sqrt{3}}{4}) \\ y = - (\frac{1}{4} + i . \frac{\sqrt{3}}{4}) \end{cases} \Rightarrow x + y \overset{?}{=} - \frac{1}{2}

x^{3} + y^{3} = (x + y) [{(x + y)}^{2} - 3 x y] =

= (- \frac{1}{2}) [\frac{1}{4} - \frac{3}{4}] = \frac{1}{4}

\Rightarrow x^{3} + y^{3} + 3 x y = \frac{1}{4} + 3 \times \frac{1}{4} = 1 o k

\Rightarrow x + y = - \frac{1}{2} [a n o t h e r p o s s i b i l i t y]]

Answered by MJS_new last updated on 24/Jul/22

x^{3} + y^{3} + 3 x y - 1 = 0

(x + y - 1) (x^{2} - x y + y^{2} + x + y + 1) = 0

(1) x + y = 1

(2) y = \frac{x - 1}{2} \pm \frac{\sqrt{3} (x + 1)}{2} i

if x, y \in R \Rightarrow x = - 1 \land y = - 1 \Rightarrow x + y = - 2

if x, y \in C \Rightarrow x + y = \frac{3 x - 1}{2} \pm \frac{\sqrt{3} (x + 1)}{2} i \land x \in C

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