Solve (a+b)^3-(a-b)^3-2b^3]^2

$$( a + b ) ^ { 3 } - ( a - b ) ^ { 3 } - 2 b ^ { 3 } ] ^ { 2 }$$

Evaluate

$$6ba^{2}$$

Solution Steps

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

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

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

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

To find the opposite of a^{3}-3a^{2}b+3ab^{2}-b^{3}, find the opposite of each term.

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

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

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

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

$$6a^{2}b+3ab^{2}+b^{3}-3ab^{2}+b^{3}-2b^{3}$$

Combine 3ab^{2} and -3ab^{2} to get 0.

$$6a^{2}b+b^{3}+b^{3}-2b^{3}$$

Combine b^{3} and b^{3} to get 2b^{3}.

$$6a^{2}b+2b^{3}-2b^{3}$$

Combine 2b^{3} and -2b^{3} to get 0.

$$6a^{2}b$$

Expand

$$6ba^{2}$$

Solution Steps

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

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

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

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

To find the opposite of a^{3}-3a^{2}b+3ab^{2}-b^{3}, find the opposite of each term.

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

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

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

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

$$6a^{2}b+3ab^{2}+b^{3}-3ab^{2}+b^{3}-2b^{3}$$

Combine 3ab^{2} and -3ab^{2} to get 0.

$$6a^{2}b+b^{3}+b^{3}-2b^{3}$$

Combine b^{3} and b^{3} to get 2b^{3}.

$$6a^{2}b+2b^{3}-2b^{3}$$

Combine 2b^{3} and -2b^{3} to get 0.

$$6a^{2}b$$