Question

[lưu tuấn anh]

Cho a + b + c + 0. Chứng minh

a^3 + b^3 + c^3 = 3abc

Answer 1

[$Mr.VôDanh$]

Ta có :

a³ +b³+c³=3abc

<=> a³+b³+c³ - 3abc =0

Áp dụng (a+b)³-3(a+b) =a³+b³

<=> (a+b)³-3(a+b) +c³-3abc=0

<=> (a+b)³+c³ -3(a+b+c)=0

<=> ( a+b+c) \(\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3\left(a+b+c\right)\)=0

<=> (a+c+b) \(\left[\left(a+b\right)^2-ca-ca+c^2-3\right]=0\)

Vì a+b+c=0 => a³+b³+c³=3abc

Answer 2

[$Mr.VôDanh$]

-ca -cb nha