Solve C=sqrt[3]{3}+(-3)^3-1+(1/8)^-3-1+(1/25)^-2-1

$$C = \sqrt[ 3 ] { 3 } + ( - 3 ) ^ { 3 - 1 } + ( \frac { 1 } { 8 } ) ^ { - 3 - 1 } + ( \frac { 1 } { 25 } ) ^ { - 2 - 1 }$$

Solve for C

$$C=\sqrt[3]{3}+19730\approx 19731.44224957$$

Steps for Solving Linear Equation

Subtract 1 from 3 to get 2.

$$C=\sqrt[3]{3}+\left(-3\right)^{2}+\left(\frac{1}{8}\right)^{-3-1}+\left(\frac{1}{25}\right)^{-2-1}$$

Calculate -3 to the power of 2 and get 9.

$$C=\sqrt[3]{3}+9+\left(\frac{1}{8}\right)^{-3-1}+\left(\frac{1}{25}\right)^{-2-1}$$

Subtract 1 from -3 to get -4.

$$C=\sqrt[3]{3}+9+\left(\frac{1}{8}\right)^{-4}+\left(\frac{1}{25}\right)^{-2-1}$$

Calculate \frac{1}{8} to the power of -4 and get 4096.

$$C=\sqrt[3]{3}+9+4096+\left(\frac{1}{25}\right)^{-2-1}$$

Add 9 and 4096 to get 4105.

$$C=\sqrt[3]{3}+4105+\left(\frac{1}{25}\right)^{-2-1}$$

Subtract 1 from -2 to get -3.

$$C=\sqrt[3]{3}+4105+\left(\frac{1}{25}\right)^{-3}$$

Calculate \frac{1}{25} to the power of -3 and get 15625.

$$C=\sqrt[3]{3}+4105+15625$$

Add 4105 and 15625 to get 19730.

$$C=\sqrt[3]{3}+19730$$

Assign C

$$C≔\sqrt[3]{3}+19730$$