Q2015/013) solve for x, y, z when

$x+y + z = - 3 \cdots 1$


and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$

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)

so $x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

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

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 = - 18- 1 + 14 = - 5

Note:

I  have  taken the problem from the site http://mathhelpboards.com/challenge-questions-puzzles-28/find-3x-y-2z-14136.html#post67082

where till this date I was the only solution provider