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Let the three real numbers be $x,y,z$. We want to find a way to relate $x^3+y^3+z^3$, $x+y+z$, and $xyz$. As a guess, we can try to expand the quantity $$\begin{array}{rl}(x+y+z{)}^3& =x^3+y^3+z^3\\ & +3x^{2}y+3x^{2}z+3y^{2}x+3y^{2}z\\ & +3z^{2}x+3z^{2}y+6xyz.\end{array}$$ Substituting $x+y+z=0$, we obtain $$x^3+y^3+z^3=-3(x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y+2xyz).$$ After some experimentation, we notice that we can group certain terms to factor out a $x+y+z$ term, such as that $x^{2}y+y^{2}x+xyz=xy(x+y+z)=0$. Similarly, $z^{2}x+x^{2}x+xyz=xz(x+y+z)$ and $y^{2}z+z^{2}y+xyz=yz(x+y+z)$. Hence, the equation reduces to $x^3+y^3+z^3=-3(-xyz)=3xyz=\boxed{51}.$

This also follows from the following identity about the sum of three cubes: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$