I will avoid expanding any of the cubed expression. Define a polynomial $p$ to be $p(x)=(x+4)^3 + (x+5)^3 - (x+7)^3 - (x-2)^3$. Note that $p$ is at most quadratic. Thus, $p(x)=ax^2+bx+c$ for some real $a,b,c$. (In fact, $a,b,c$ are integers, but it's enough to say they are real.) We have
\begin{eqnarray*}
p(0) = c &=& 4^3+5^3-7^3 - (-2)^3 = -146 \\
p(1) = a+b+c &=& 5^3+6^3-8^3-(-1)^3 = -170 \\
p(-1) = a-b+c &=& 3^3+4^3-6^3-(-3)^3 = -98
\end{eqnarray*}
Solving for $a$, $b$, and $c$, we have $a=12$, $b=-36$, and $c=-146$, giving $p(x)= 12x^2-36x-146$. So
\[
x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{3}{2}\pm\dfrac{1}{6\sqrt{519}}.
\]
idk if this is right
\(x=\frac{9+\sqrt{519}}{6},\:x=\frac{9-\sqrt{519}}{6}\)