A) \( \Large a^{2} \left(b+c\right)+b^{2} \left(c+a\right)+c^{2} \left(a+b\right) \) |
B) \( \Large 3 \left(b+c\right) \left(c+a\right) \left(a+b\right) \) |
C) \( \Large 3 abc \) |
D) \( \Large 6 a^{2}b^{2}c^{2} \) |
C) \( \Large 3 abc \) |
Given: \( \Large a+b+c=0 \)
=> \( \Large a+b = -c \)
Taking cube on both sides, we have
\( \Large \left(a+b\right)^{3} = \left(-c\right)^{3} \)
=> \( \Large a^{3}+b^{3}+3ab \left(a+b\right)=-c^{3} \)
=> \( \Large a^{3}+b^{3}+3ab \left(-c\right)=-c^{3} \)
=> \( \Large a^{3}+b^{3}+c^{3}=3abc \)