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

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

Evaluate

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

Solution Steps

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

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

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

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

Combine 8a^{3} and a^{3} to get 9a^{3}.

$$9a^{3}+12a^{2}+6a+1-3a^{2}+3a-1$$

Combine 12a^{2} and -3a^{2} to get 9a^{2}.

$$9a^{3}+9a^{2}+6a+1+3a-1$$

Combine 6a and 3a to get 9a.

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

Subtract 1 from 1 to get 0.

$$9a^{3}+9a^{2}+9a$$

Expand

$$9a^{3}+9a^{2}+9a$$

Solution Steps

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

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

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

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

Combine 8a^{3} and a^{3} to get 9a^{3}.

$$9a^{3}+12a^{2}+6a+1-3a^{2}+3a-1$$

Combine 12a^{2} and -3a^{2} to get 9a^{2}.

$$9a^{3}+9a^{2}+6a+1+3a-1$$

Combine 6a and 3a to get 9a.

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

Subtract 1 from 1 to get 0.

$$9a^{3}+9a^{2}+9a$$