Solve (3^{-2}x^-3)+(2/3)^-3-13^0*(-1)^2

$$( 3 ^ { - 2 } x ^ { - 3 } ) + ( \frac { 2 } { 3 } ) ^ { - 3 } - 13 ^ { 0 } \times ( - 1 ) ^ { 2 }$$

Evaluate

$$\frac{19}{8}+\frac{1}{9x^{3}}$$

Solution Steps

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$\frac{1}{9}x^{-3}+\left(\frac{2}{3}\right)^{-3}-13^{0}\left(-1\right)^{2}$$

Calculate \frac{2}{3} to the power of -3 and get \frac{27}{8}.

$$\frac{1}{9}x^{-3}+\frac{27}{8}-13^{0}\left(-1\right)^{2}$$

Calculate 13 to the power of 0 and get 1.

$$\frac{1}{9}x^{-3}+\frac{27}{8}-1\left(-1\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$\frac{1}{9}x^{-3}+\frac{27}{8}-1\times 1$$

Multiply 1 and 1 to get 1.

$$\frac{1}{9}x^{-3}+\frac{27}{8}-1$$

Subtract 1 from \frac{27}{8} to get \frac{19}{8}.

$$\frac{1}{9}x^{-3}+\frac{19}{8}$$

Factor

$$\frac{171x^{3}+8}{72x^{3}}$$