MHB Solving for $3x+y+2z$ Given Integer Constraints

[anemone]

If $x,\,y,\,z \in Z$ and that $z\ge y \ge x$ and also,

$x+y+z=-3$

$x^3+y^3+z^3-20(x+3)(y+3)(z+3)=2013$

evaluate the value for $3x+y+2z$.

 

[Albert1]

If $x,\,y,\,z \in N$ and that $z\ge y \ge x$ and also,

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

$x^3+y^3+z^3-20(x+3)(y+3)(z+3)=2013$

evaluate the value for $3x+y+2z$.

$x,y,z\in N,$
why $x+y+z=-3 ? $ in (1)

 

[anemone]

$x,y,z\in N,$
why $x+y+z=-3 ? $ in (1)

It was a mistake, sorry, Albert, you are right, the three of them are integers, and thanks for catching!

 

[kaliprasad]

Spoiler

We have
$x+y + z = - 3 \cdots 1$
and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$ from (1)
$x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

Now
$(x+y+z)^3 = x^3+y^3+z^+3(x+y)(y+z)(z+x)$

or $-27 = 2013 + 20(x+3)(y+3)(z+3) - 3(z+3)(x+3)(y+3)$ (LHS from (1) and RHS from (2), (3),(4),(5)

so $ 17(x+3)(y+3)(z+3) = -(2013+27)=-2040$

or $(x+3)(y+3)(z+3)= - 120$
further $x + 3 + y +3 + z + 3 = 6$
so we need 3 integers product is -120 and sum 6 and the numbers are 10,2,-6
so z = 7, y = -1, x = - 9
$3x + y + 2z = - 27- 1 + 14 = - 14$

 

[anemone]

Spoiler

We have
$x+y + z = - 3 \cdots 1$
and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$ from (1)
$x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

Now
$(x+y+z)^3 = x^3+y^3+z^+3(x+y)(y+z)(z+x)$

or $-27 = 2013 + 20(x+3)(y+3)(z+3) - 3(z+3)(x+3)(y+3)$ (LHS from (1) and RHS from (2), (3),(4),(5)

so $ 17(x+3)(y+3)(z+3) = -(2013+27)=-2040$

or $(x+3)(y+3)(z+3)= - 120$
further $x + 3 + y +3 + z + 3 = 6$
so we need 3 integers product is -120 and sum 6 and the numbers are 10,2,-6
so z = 7, y = -1, x = - 9
$3x + y + 2z = - 27- 1 + 14 = - 14$

Very well done, kaliprasad!