Solve for y
y=\left(x-12\right)^{3}-10
Solve for x
x=\sqrt[3]{y+10}+12
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y=x^{3}-36x^{2}+432x-1728-10
Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-12\right)^{3}.
y=x^{3}-36x^{2}+432x-1738
Subtract 10 from -1728 to get -1738.
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