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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Chứng minh rằng `:` Nếu `a+b+c = 0` thì `a^3 + b^3 + c^3 = 3abc`

Toán Lớp 8: Chứng minh rằng : Nếu a+b+c = 0 thì a^3 + b^3 + c^3 = 3abc

Comments ( 2 )

  1. Ta có: a+b+c=0
    ->(a+b+c)^3=0
    ->[(a+b)+c]^3=0
    ->(a+b)^3+3(a+b)^2 .c+3(a+b).c^2+c^3=0
    ->a^3+3a^2b+3ab^2+b^3+3.(a^2+2ab+b^2).c+3ac^2_3bc^2+c^3=0
    ->a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc=0
    ->a^3+b^3+c^3+3a^2b+3abc^2+3a^2c+3ac^2_3b^2c+3bc^2+9abc=3abc
    ->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+(a+b+c).(3ab+3bc+3ac)=3abc
    ->a^3+b^3+c^3+0.(3ab+3bc+3ac)=3abc
    ->a^3+b^3+c^3=3abc(đpcm)

  2. Giải đáp+Lời giải và giải thích chi tiết:
     Ta có: a+b+c=0
    =>a+b=-c
    $+)a^3 + b^3 + c^3 \\ =(a+b)(a^2-ab+b^2)+c^2\\=-c(a^2+2ab+b^2-3ab)+c^2\\=-c(a+b)^2+3abc+c^2\\=3abc$

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222-9+11+12:2*14+14 = ? ( )

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