$\text{Cho a+b+c=0}$ $\text{Cmr: a^3+b^3+c^3=3abc}$ 03/07/2021 Bởi Gianna $\text{Cho a+b+c=0}$ $\text{Cmr: a^3+b^3+c^3=3abc}$
Ta có: `a+b+c=0` `<=> (a+b+c)^3=0` `<=> a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2+6abc=0` `<=> ( a^3+b^3+c^3)+(3a^2b+3ab^2+3abc)+(3b^2c+3bc^2+3abc)+(3c^2a+3ca^2+3abc)-3abc=0` `<=> (a^3+b^3+c^3)+3ab(a+b+c) + 3bc(a+b+c)+3ac(a+b+c)-3abc` `<=> a^3+b^3+c^3=3abc` Bình luận
Ta có: ` a^3+b^3=(a + b)^3-3ab(a + b)` Nên `a^3+b^3+c^3=(a + b)^3-3ab(a + b)+c^3(1)` Mà `a+b+c=0⇒a+b =-c(2)` Thay `(2)` vào `(1)` ta được: `a^3+b^3+c^3=(-c)^3-3ab(-c)+c^3 =-c^3+3abc+c^3=3abc(đpcm)` Bình luận
Ta có: `a+b+c=0`
`<=> (a+b+c)^3=0`
`<=> a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2+6abc=0`
`<=> ( a^3+b^3+c^3)+(3a^2b+3ab^2+3abc)+(3b^2c+3bc^2+3abc)+(3c^2a+3ca^2+3abc)-3abc=0`
`<=> (a^3+b^3+c^3)+3ab(a+b+c) + 3bc(a+b+c)+3ac(a+b+c)-3abc`
`<=> a^3+b^3+c^3=3abc`
Ta có: ` a^3+b^3=(a + b)^3-3ab(a + b)`
Nên `a^3+b^3+c^3=(a + b)^3-3ab(a + b)+c^3(1)`
Mà `a+b+c=0⇒a+b =-c(2)`
Thay `(2)` vào `(1)` ta được:
`a^3+b^3+c^3=(-c)^3-3ab(-c)+c^3 =-c^3+3abc+c^3=3abc(đpcm)`