Solve (2*8)^{2}-(625)^0+(8*8)^3+3^2/3^3

$$( 2 \times 8 ) ^ { 2 } - ( 625 ) ^ { 0 } + ( 8 \times 8 ) ^ { 3 } + \frac { 3 ^ { 2 } } { 3 ^ { 3 } }$$

Evaluate

$$\frac{787198}{3}\approx 262399.333333333$$

Solution Steps

Rewrite 3^{3} as 3\times 3^{2}. Cancel out 3^{2} in both numerator and denominator.

$$\left(2\times 8\right)^{2}-625^{0}+\left(8\times 8\right)^{3}+\frac{1}{3}$$

Multiply 2 and 8 to get 16.

$$16^{2}-625^{0}+\left(8\times 8\right)^{3}+\frac{1}{3}$$

Calculate 16 to the power of 2 and get 256.

$$256-625^{0}+\left(8\times 8\right)^{3}+\frac{1}{3}$$

Calculate 625 to the power of 0 and get 1.

$$256-1+\left(8\times 8\right)^{3}+\frac{1}{3}$$

Subtract 1 from 256 to get 255.

$$255+\left(8\times 8\right)^{3}+\frac{1}{3}$$

Multiply 8 and 8 to get 64.

$$255+64^{3}+\frac{1}{3}$$

Calculate 64 to the power of 3 and get 262144.

$$255+262144+\frac{1}{3}$$

Add 255 and 262144 to get 262399.

$$262399+\frac{1}{3}$$

Add 262399 and \frac{1}{3} to get \frac{787198}{3}.

$$\frac{787198}{3}$$

Factor

$$\frac{2 \cdot 23 \cdot 109 \cdot 157}{3} = 262399\frac{1}{3} = 262399.3333333333$$