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

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

Evaluate

$$9\left(a^{2}-a+1\right)$$

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}.

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

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

$$a^{3}+3a^{2}+3a+1-\left(a^{3}-6a^{2}+12a-8\right)$$

To find the opposite of a^{3}-6a^{2}+12a-8, find the opposite of each term.

$$a^{3}+3a^{2}+3a+1-a^{3}+6a^{2}-12a+8$$

Combine a^{3} and -a^{3} to get 0.

$$3a^{2}+3a+1+6a^{2}-12a+8$$

Combine 3a^{2} and 6a^{2} to get 9a^{2}.

$$9a^{2}+3a+1-12a+8$$

Combine 3a and -12a to get -9a.

$$9a^{2}-9a+1+8$$

Add 1 and 8 to get 9.

$$9a^{2}-9a+9$$

Expand

$$9a^{2}-9a+9$$

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}.

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

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

$$a^{3}+3a^{2}+3a+1-\left(a^{3}-6a^{2}+12a-8\right)$$

To find the opposite of a^{3}-6a^{2}+12a-8, find the opposite of each term.

$$a^{3}+3a^{2}+3a+1-a^{3}+6a^{2}-12a+8$$

Combine a^{3} and -a^{3} to get 0.

$$3a^{2}+3a+1+6a^{2}-12a+8$$

Combine 3a^{2} and 6a^{2} to get 9a^{2}.

$$9a^{2}+3a+1-12a+8$$

Combine 3a and -12a to get -9a.

$$9a^{2}-9a+1+8$$

Add 1 and 8 to get 9.

$$9a^{2}-9a+9$$