Replacing $x$ is not a good way.

Here's a solution WITHOUT plugging in values for x :

Finding integer solutions to the expression is easier when it's factored. So I'll turn this into a cubic expression & rearrange a bit: $x^3 + x^2 + x - 39 = 0 \\ x^3 - 39 + x^2 + x = 0 \\$ I'd like to pair up $x^3$ with another fellow cube (for easier factoring), so I'll split up $39$ into $27$ & $12$ and rearrange a bit more:

$x^3 - 27 + x^2 + x - 12 = 0 \\ \Rightarrow (x-3)(x^2 + 3x + 9) + (x-3)(x+4) = 0 \\ \Rightarrow (x-3)(x^2 + 3x + 9 + x + 4) = 0 \\ \Rightarrow (x-3)(x^2 + 4x + 13) = 0$

As we can see, the second term doesn't have any integer solutions, so $x$ has to be $\boxed{3}$ . No plugging in necessary :D

$1^3=1, ~~ 2^3=8,~~ 3^3=27, 4^3=64. ~~~\\Since~~in~~ x + x^2 + x^3 = 39.~~\\39 ~is~ a ~posative~integer~\implies~ ~x~must~be~a~posative~integer. \\x+x^2\geq0.~~~~~~~~~~~27<39<64~~~~~~~~~~\therefore~3\leq x<4.\\On~ checking~x=~\boxed{ 3 }$

its very easy wee have to put x=3 to get our ans. equals to 39