Solve (a-1)^3-(a-1)^2-2(a-1)

$$( a - 1 ) ^ { 3 } - ( a - 1 ) ^ { 2 } - 2 ( a - 1 )$$

Evaluate

$$a\left(a-3\right)\left(a-1\right)$$

Solution Steps

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a-1\right)^{3}.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-1\right)^{2}.

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

To find the opposite of a^{2}-2a+1, find the opposite of each term.

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

Combine -3a^{2} and -a^{2} to get -4a^{2}.

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

Combine 3a and 2a to get 5a.

$$a^{3}-4a^{2}+5a-1-1-2\left(a-1\right)$$

Subtract 1 from -1 to get -2.

$$a^{3}-4a^{2}+5a-2-2\left(a-1\right)$$

Use the distributive property to multiply -2 by a-1.

$$a^{3}-4a^{2}+5a-2-2a+2$$

Combine 5a and -2a to get 3a.

$$a^{3}-4a^{2}+3a-2+2$$

Add -2 and 2 to get 0.

$$a^{3}-4a^{2}+3a$$

Expand

$$a^{3}-4a^{2}+3a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a-1\right)^{3}.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-1\right)^{2}.

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

To find the opposite of a^{2}-2a+1, find the opposite of each term.

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

Combine -3a^{2} and -a^{2} to get -4a^{2}.

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

Combine 3a and 2a to get 5a.

$$a^{3}-4a^{2}+5a-1-1-2\left(a-1\right)$$

Subtract 1 from -1 to get -2.

$$a^{3}-4a^{2}+5a-2-2\left(a-1\right)$$

Use the distributive property to multiply -2 by a-1.

$$a^{3}-4a^{2}+5a-2-2a+2$$

Combine 5a and -2a to get 3a.

$$a^{3}-4a^{2}+3a-2+2$$

Add -2 and 2 to get 0.

$$a^{3}-4a^{2}+3a$$