Solve for x
x=\frac{\left(y+4\right)^{3}}{3}
Solve for y
y=\sqrt[3]{3x}-4
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x=\frac{y^{3}+12y^{2}+48y+64}{3}
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(y+4\right)^{3}.
x=\frac{1}{3}y^{3}+4y^{2}+16y+\frac{64}{3}
Divide each term of y^{3}+12y^{2}+48y+64 by 3 to get \frac{1}{3}y^{3}+4y^{2}+16y+\frac{64}{3}.
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