Solve {[(a^3-1):(a-1)-a-2]^3:(a+1)^3+1}:a

$$\{ [ ( a ^ { 3 } - 1 ) : ( a - 1 ) - a - 2 ] ^ { 3 } : ( a + 1 ) ^ { 3 } + 1 \} : a$$

Evaluate

$$a^{2}-3a+3$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{3}-1}{a-1}.

$$\frac{\frac{\left(\frac{\left(a-1\right)\left(a^{2}+a+1\right)}{a-1}-a-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Cancel out a-1 in both numerator and denominator.

$$\frac{\frac{\left(a^{2}+a+1-a-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Combine a and -a to get 0.

$$\frac{\frac{\left(a^{2}+1-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Subtract 2 from 1 to get -1.

$$\frac{\frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}}.

$$\frac{\frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}}+\frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Since \frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}} and \frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(a^{2}-1\right)^{3}+\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Do the multiplications in \left(a^{2}-1\right)^{3}+\left(a+1\right)^{3}.

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

Combine like terms in \left(a^{2}\right)^{3}+3\left(a^{2}\right)^{2}\left(-1\right)+3a^{2}-1+a^{3}+3a^{2}+3a\times 1^{2}+1^{3}.

$$\frac{\frac{3a+a^{6}-3a^{4}+6a^{2}+a^{3}}{\left(a+1\right)^{3}}}{a}$$

Factor the expressions that are not already factored in \frac{3a+a^{6}-3a^{4}+6a^{2}+a^{3}}{\left(a+1\right)^{3}}.

$$\frac{\frac{a\left(a^{2}-3a+3\right)\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Cancel out \left(a+1\right)^{3} in both numerator and denominator.

$$\frac{a\left(a^{2}-3a+3\right)}{a}$$

Cancel out a in both numerator and denominator.

$$a^{2}-3a+3$$

Expand

$$a^{2}-3a+3$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{3}-1}{a-1}.

$$\frac{\frac{\left(\frac{\left(a-1\right)\left(a^{2}+a+1\right)}{a-1}-a-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Cancel out a-1 in both numerator and denominator.

$$\frac{\frac{\left(a^{2}+a+1-a-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Combine a and -a to get 0.

$$\frac{\frac{\left(a^{2}+1-2\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

Subtract 2 from 1 to get -1.

$$\frac{\frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}}+1}{a}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}}.

$$\frac{\frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}}+\frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Since \frac{\left(a^{2}-1\right)^{3}}{\left(a+1\right)^{3}} and \frac{\left(a+1\right)^{3}}{\left(a+1\right)^{3}} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(a^{2}-1\right)^{3}+\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Do the multiplications in \left(a^{2}-1\right)^{3}+\left(a+1\right)^{3}.

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

Combine like terms in \left(a^{2}\right)^{3}+3\left(a^{2}\right)^{2}\left(-1\right)+3a^{2}-1+a^{3}+3a^{2}+3a\times 1^{2}+1^{3}.

$$\frac{\frac{3a+a^{6}-3a^{4}+6a^{2}+a^{3}}{\left(a+1\right)^{3}}}{a}$$

Factor the expressions that are not already factored in \frac{3a+a^{6}-3a^{4}+6a^{2}+a^{3}}{\left(a+1\right)^{3}}.

$$\frac{\frac{a\left(a^{2}-3a+3\right)\left(a+1\right)^{3}}{\left(a+1\right)^{3}}}{a}$$

Cancel out \left(a+1\right)^{3} in both numerator and denominator.

$$\frac{a\left(a^{2}-3a+3\right)}{a}$$

Cancel out a in both numerator and denominator.

$$a^{2}-3a+3$$