# Question b8484

Sep 20, 2016

$\left(\begin{matrix}x = - 1 & y = - 1 \\ x = 5 - \sqrt{6} & y = \frac{1}{3} \left(3 - \sqrt{6}\right) \\ x = 5 + \sqrt{6} & y = \frac{1}{3} \left(3 + \sqrt{6}\right)\end{matrix}\right)$

#### Explanation:

Firstly, you should be more careful when copying and publishing problems. The correct post should be:

{(x^3+1 = 81(y^2+y)) , (x^2+x= 9(y^3 +1)):}

because so, adding term to term

{(x^3+1 = 3 xx 27(y^2+y)) , (3(x^2+x)= 27(y^3 +1)):}

we obtain

${\left(x + 1\right)}^{3} = 27 {\left(y + 1\right)}^{3}$

and

$x + 1 = 3 \left(y + 1\right)$

Now solving the system

$\left\{\begin{matrix}x + 1 = 3 \left(y + 1\right) \\ {x}^{2} + x = 9 \left({y}^{3} + 1\right)\end{matrix}\right.$

we obtain

$\left(\begin{matrix}x = - 1 & y = - 1 \\ x = 5 - \sqrt{6} & y = \frac{1}{3} \left(3 - \sqrt{6}\right) \\ x = 5 + \sqrt{6} & y = \frac{1}{3} \left(3 + \sqrt{6}\right)\end{matrix}\right)$

Note:

To solve

$\left\{\begin{matrix}x + 1 = 3 \left(y + 1\right) \\ {x}^{2} + x = 9 \left({y}^{3} + 1\right)\end{matrix}\right.$ we proceed as follows:

In the second equation

x^2+x=x(x+1) = x(3(y+1))=9 (y^3 + 1))# then

$x = \frac{9}{3} \frac{{y}^{3} + 1}{y + 1} = 3 \left(1 - y + {y}^{2}\right)$ and finally

$x = 3 y + 3 - 1 = 3 - 3 y + 3 {y}^{2}$ or

$3 {y}^{2} - 6 y + 1 = 0$