Solve for y
y=\left(x+3\right)^{3}-3
Solve for x
x=\sqrt[3]{y+3}-3
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y=x^{3}+9x^{2}+27x+27-3
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+3\right)^{3}.
y=x^{3}+9x^{2}+27x+24
Subtract 3 from 27 to get 24.
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