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Ta có : \(x=a^2-bc\Rightarrow ax=a^3-abc\); \(y=b^2-ac\Rightarrow by=b^3-abc\); \(z=c^2-ab\Rightarrow cz=c^3-abc\)
\(\Rightarrow ax+by+cz=a^3+b^3+c^3-3abc\)
Ta có : \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2-bc+b^2-ac+c^2-ab\right)=\left(a+b+c\right)\left(x+y+z\right)\)
Vậy : \(\left(x+y+z\right)\left(a+b+c\right)=ax+by+cz\)(đpcm)
Bạn lưu ý đề bài ở chỗ \(y^2=b^2-ac\)bạn ghi sai nhé, phải là \(y=b^2-ac\)
Bạn nhớ ghi thêm điều kiện x,y,z khác 0 nữa nhé :))
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b) Ta có: \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)
\(=\left(ab^2-cb^2\right)+\left(ca^2-c^2a\right)+\left(bc^2-ba^2\right)\)
\(=b^2\left(a-c\right)+ca\left(a-c\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2+ca\right)-b\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(b^2+ca-ba-bc\right)\)
\(=\left(a-c\right)\left[b\left(b-a\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left(b-a\right)\left(b-c\right)\)
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Lời giải:
Thực hiện khai triển ta có:
\((x+y+z)(a+b+c)=ax+by+xz+x(b+c)+y(a+c)+z(a+b)\)
\(=ax+by+cz+(a^2-bc)(b+c)+(b^2-ac)(a+c)+(c^2-ab)(a+b)\)
\(=ax+by+cz+(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)-(b^2c+bc^2+a^2c+ac^2+a^2b+ab^2)\)
\(=ax+by+cz+(a^2b-a^2b)+(ab^2-ab^2)+(b^2c-b^2c)+(bc^2-bc^2)+(ac^2-ac^2)+(a^2c-a^2c)\)
\(=ax+by+cz\)
Ta có đpcm.