Question

a, Cmr : ( a + b + c ). ( a^2 + b^2 + c^2 -ab - ac - bc ) = a^3 + b^3 + c^3 -3abc 

b, Áp dụng :

a+b+c= 0 thì

a^3 + b^3 + c^3  = 3abc 

Answer 1

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

b,

Ta có:

\(\left(a+b+c\right)^3=0\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)