Solve (0.362)^3-(0.1)^3/(0.362)^2+0.0362+(0.1)^2

$$\frac { ( 0.362 ) ^ { 3 } - ( 0.1 ) ^ { 3 } } { ( 0.362 ) ^ { 2 } + 0.0362 + ( 0.1 ) ^ { 2 } }$$

Evaluate

$$0.262$$

Solution Steps

Calculate 0.362 to the power of 3 and get 0.047437928.

$$\frac{0.047437928-0.1^{3}}{0.362^{2}+0.0362+0.1^{2}}$$

Calculate 0.1 to the power of 3 and get 0.001.

$$\frac{0.047437928-0.001}{0.362^{2}+0.0362+0.1^{2}}$$

Subtract 0.001 from 0.047437928 to get 0.046437928.

$$\frac{0.046437928}{0.362^{2}+0.0362+0.1^{2}}$$

Calculate 0.362 to the power of 2 and get 0.131044.

$$\frac{0.046437928}{0.131044+0.0362+0.1^{2}}$$

Add 0.131044 and 0.0362 to get 0.167244.

$$\frac{0.046437928}{0.167244+0.1^{2}}$$

Calculate 0.1 to the power of 2 and get 0.01.

$$\frac{0.046437928}{0.167244+0.01}$$

Add 0.167244 and 0.01 to get 0.177244.

$$\frac{0.046437928}{0.177244}$$

Expand \frac{0.046437928}{0.177244} by multiplying both numerator and the denominator by 1000000000.

$$\frac{46437928}{177244000}$$

Reduce the fraction \frac{46437928}{177244000} to lowest terms by extracting and canceling out 354488.

$$\frac{131}{500}$$

Factor

$$\frac{131}{2 ^ {2} \cdot 5 ^ {3}} = 0.262$$