Determine-the-value-of-x-y-if-x-3-y-3-1-x-y-x-1-y-1-2-

Question Number 98104 by bemath last updated on 11/Jun/20

Determine the value of x + y

if {\begin{cases} x^{3} + y^{3} = 1 \\ (x + y) (x + 1) (y + 1) = 2 \end{cases}

Commented by bemath last updated on 12/Jun/20

thank you all

Answered by mr W last updated on 11/Jun/20

u = x + y

v = x y

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

\Rightarrow u^{3} - 3 u v = 1

\Rightarrow v = \frac{u^{3} - 1}{3 u}

(x + y) (x y + x + y + 1) = 2

\Rightarrow u (u + v + 1) = 2

\Rightarrow u (u + \frac{u^{3} - 1}{3 u} + 1) = 2

\Rightarrow u^{3} + 3 u^{2} + 3 u - 1 = 6

\Rightarrow u^{3} + 3 u^{2} + 3 u + 1 = 8

\Rightarrow {(u + 1)}^{3} = 8 = 2^{3}

\Rightarrow u + 1 = 2, 2 ω, 2 ω^{2}

\Rightarrow u = x + y = 1 (r e a l)

\Rightarrow u = x + y = 2 ω - 1 = - 2 + i \sqrt{3}

\Rightarrow u = x + y = 2 ω^{2} - 1 = - 2 - i \sqrt{3}

Commented by Lekhraj last updated on 11/Jun/20

{(u + 1)}^{3} = 8

Commented by mr W last updated on 11/Jun/20

y e s

Answered by Rasheed.Sindhi last updated on 11/Jun/20

if {\begin{cases} x^{3} + y^{3} = 1 \dots \dots \dots \dots \dots \dots \dots . . (i) \\ (x + y) (x + 1) (y + 1) = 2 \dots . . (ii) \end{cases}

(ii) \Rightarrow

xy = \frac{2}{x + y} - (x + y) - 1 \dots . . (iii)

(i) \Rightarrow

{(x + y)}^{3} - 3 xy (x + y) = 1 \dots \dots (iv)

(iii) & (iv) \Rightarrow

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

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

{(x + y)}^{3} - 6 + 3 {(x + y)}^{2} + 3 (x + y)) = 1

{(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) - 7 = 0

t^{3} + {3 t}^{2} + 3 t - 7 = 0

- 7 = (\pm 1) (\mp 7)

t = 1 satisfy the equation

x + y = 1

∴ t - 1 is factor of t^{3} + {3 t}^{2} + 3 t - 7

(\begin{matrix} 1) & 1 & 3 & 3 & - 7 \\ 1 & 4 & 7 \\ 1 & 4 & 7 & 0 \end{matrix})

t^{2} + 4 t + 7 = 0

t = \frac{- 4 \pm \sqrt{16 - 28}}{2}

t = \frac{- 4 \pm 2 i \sqrt{3}}{2}

x + y = \frac{- 4 \pm 2 i \sqrt{3}}{2} = - 2 \pm i \sqrt{3}

x + y = 1, - 2 \pm i \sqrt{3}

Answered by 1549442205 last updated on 11/Jun/20

(2) \Leftrightarrow (x + y) [xy + (x + y) + 1] = 2

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

Put x + y = u, xy = v we have {\begin{cases} u^{3} - 3 uv = 1 (3) \\ u (u + v + 1) = 2 (4) \end{cases}

From (4) we ger uv = 2 - u^{2} - u . Replace into (3) we get

u^{3} - 3 (2 - u^{2} - u) = 1 \Leftrightarrow u^{3} + {3 u}^{2} + 3 u - 7 = 9

\Leftrightarrow (u - 1) (u^{2} + 4 u + 7) = 0 \Leftrightarrow u - 1 = 0 (as u^{2} + 4 u + 7 = {(u + 2)}^{2} + 3 > 0)

, so u = 1, replace into (4) we get v = 0 . We have

{\begin{cases} x + y = 1 \\ xy = 0 \end{cases} \Leftrightarrow (x; y) \in {(0; 1); (1; 0)}

Thus, our system of equations has two

soluions : (x; y) \in {(0; 1); (1; 0)}

Answered by MJS last updated on 11/Jun/20

x = u - v \land y = u + v

{\begin{cases} 2 u^{3} + 6 {u v}^{2} - 1 = 0 \\ 2 (u^{3} + 2 u^{2} + u - 1) - 2 {u v}^{2} = 0 \end{cases}

\Rightarrow

v^{2} = \frac{1 - 2 u^{3}}{6 u} = \frac{u^{3} + 2 u^{2} + u - 1}{u}

\Rightarrow

u^{3} + \frac{3}{2} u^{2} + \frac{3}{4} u - \frac{7}{8} = 0

u = t - \frac{1}{2}

t^{3} - 1 = 0

t = 1 \lor t = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i

u = \frac{1}{2} \lor u = - 1 \pm \frac{\sqrt{3}}{2} i

x = u - v \land y = u + v \Rightarrow x + y = 2 u

answer is 1 \lor - 2 \pm \sqrt{3} i

Answered by maths mind last updated on 11/Jun/20

(x + y) (x + 1) (y + 1)

= (x + y) (x y + x + y + 1)

= {x y (x + y) + {(x + y)}^{2} + (x + y) = 2} \times 3 . . E

x^{3} + y^{3} = 1 . . G

E + G = {(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) = 7

\Leftrightarrow {(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) + 1 = 8

\Leftrightarrow {(1 + x + y)}^{3} = 8 \Rightarrow 1 + x + y = 2 e^{i \frac{2 k π}{3}}

\Rightarrow x + y = 2 e^{i \frac{k π}{3}} - 1, k \in {0, 1, 2}