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

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

Evaluate

$$4a$$

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}.

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

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

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-\left(2a\right)^{2}-1$$

Expand \left(2a\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-2^{2}a^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-4a^{2}-1$$

Use the distributive property to multiply -2a by 4a^{2}+4a+1.

$$8a^{3}+12a^{2}+6a+1-8a^{3}-8a^{2}-2a-4a^{2}-1$$

Combine 8a^{3} and -8a^{3} to get 0.

$$12a^{2}+6a+1-8a^{2}-2a-4a^{2}-1$$

Combine 12a^{2} and -8a^{2} to get 4a^{2}.

$$4a^{2}+6a+1-2a-4a^{2}-1$$

Combine 6a and -2a to get 4a.

$$4a^{2}+4a+1-4a^{2}-1$$

Combine 4a^{2} and -4a^{2} to get 0.

$$4a+1-1$$

Subtract 1 from 1 to get 0.

$$4a$$

Expand

$$4a$$

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}.

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

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

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-\left(2a\right)^{2}-1$$

Expand \left(2a\right)^{2}.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-2^{2}a^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$8a^{3}+12a^{2}+6a+1-2a\left(4a^{2}+4a+1\right)-4a^{2}-1$$

Use the distributive property to multiply -2a by 4a^{2}+4a+1.

$$8a^{3}+12a^{2}+6a+1-8a^{3}-8a^{2}-2a-4a^{2}-1$$

Combine 8a^{3} and -8a^{3} to get 0.

$$12a^{2}+6a+1-8a^{2}-2a-4a^{2}-1$$

Combine 12a^{2} and -8a^{2} to get 4a^{2}.

$$4a^{2}+6a+1-2a-4a^{2}-1$$

Combine 6a and -2a to get 4a.

$$4a^{2}+4a+1-4a^{2}-1$$

Combine 4a^{2} and -4a^{2} to get 0.

$$4a+1-1$$

Subtract 1 from 1 to get 0.

$$4a$$