Solve n^2+45478+56747*352/9^2+b^3=c

$$n ^ { 2 } + 45478 + 56747 \times \frac { 352 } { 9 ^ { 2 } } + b ^ { 3 } = c$$

Solve for c

$$c=n^{2}+b^{3}+\frac{23658662}{81}$$

Solution Steps

Calculate 9 to the power of 2 and get 81.

$$n^{2}+45478+56747\times \frac{352}{81}+b^{3}=c$$

Multiply 56747 and \frac{352}{81} to get \frac{19974944}{81}.

$$n^{2}+45478+\frac{19974944}{81}+b^{3}=c$$

Add 45478 and \frac{19974944}{81} to get \frac{23658662}{81}.

$$n^{2}+\frac{23658662}{81}+b^{3}=c$$

Swap sides so that all variable terms are on the left hand side.

$$c=n^{2}+\frac{23658662}{81}+b^{3}$$

Solve for b

$$b=\frac{3^{\frac{2}{3}}\sqrt[3]{-81n^{2}+81c-23658662}}{9}$$