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Let $x=\frac{a}{b},$ $y=\frac{b}{c},$ and $z=\frac{c}{a}.$ Then $x+y+z=7$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=9.$ Also, $$xyz=\frac{a}{b}\cdot \frac{b}{c}\cdot \frac{c}{a}=1,$$so $xy+xz+yz=9.$

We want to compute $x^3+y^3+z^3.$ Recall the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$Squaring the equation $x+y+z=7,$ we get $$x^2+y^2+z^2+2(xy+xz+yz)=49.$$Then $$x^2+y^2+z^2-xy-xz-yz=49-3(xy+xz+yz)=49-3\cdot 9=22.$$Hence, $$x^3+y^3+z^3=7\cdot 22+3=\boxed{157}.$$