Solve: (a+b)3−(a−b)3.
We know, (a+b)3=a3+b3+3a2b+3ab2
⟹ (a+b)3=a3+b3+3ab(a+b).
Also, (a−b)3=a3−b3−3a2b+3ab2
⟹ (a−b)3=a3−b3−3ab(a−b).
Then, (a+b)3− (a−b)3
=a3+b3+3a2b+3ab2 −(a3−b3−3a2b+3ab2)
=a3+b3+3a2b+3ab2 −a3+b3+3a2b−3ab2
=b3+3a2b +b3+3a2b
=2b3+6a2b
=2b(b2+3a2).
Therefore, option B is correct.