Question #b8484

Question

2084 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-20T11:38:34

$((x=-1,y=-1),(x=5-sqrt(6),y=1/3(3-sqrt(6))),(x=5+sqrt(6),y=1/3(3+sqrt(6))))$

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

$(x+1)^3=27(y+1)^3$

and

$x+1=3(y+1)$

Now solving the system

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

we obtain

$((x=-1,y=-1),(x=5-sqrt(6),y=1/3(3-sqrt(6))),(x=5+sqrt(6),y=1/3(3+sqrt(6))))$

Note:

To solve

${(x+1=3(y+1)),(x^2 + x = 9 (y^3 + 1)):}$ we proceed as follows:

In the second equation

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

$x = 9/3(y^3+1)/(y+1) = 3(1-y+y^2)$ and finally

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

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