Question

[Tiếng anh123456]

Cho a+b+c=0 . CM : a^3 + b^3 + a^2.c + b^2.c - abc = 0

Answer 1

[Gấuu]

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

Chứng minh: \(VP=\left(x+y\right)\left(x^2-xy+y^2\right)=x^3-x^2y+xy^2+x^2y-xy^2+y^3=x^3+y^3=VP\)

Áp dụng vào bài 

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Ta có \(a+b+c=0\Leftrightarrow-c=a+b\)

\(\Rightarrow c^2=\left(a+b\right)\left(a+b\right)=a^2+2ab+b^2\)

Xét \(a^3+b^3+a^2c+b^2c-abc\)

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

\(=a^3+b^3+c.c^2-3abc\)

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

\(=a^3+a^2b+2a^2b+2ab^2+ab^2+b^3-3a^2b-3ab^2+c^3-3abc\)

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

\(=\left(a+b\right)\left(a^2+2ab+b^2\right)+c^3\) ( do a+b+c=0 )

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

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

( Áp dụng \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\) )

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

Vậy \(a^3+b^3+a^2c+b^2c-abc=0\)