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2
a
\(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)
b
Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)
Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)
Áp dụng kết quả câu a ta được:
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
a)(a+b+c)3 - a3 - b3 - c3
= (a+b+c-a)( a2+b2+c2+2ab+2bc+2ac-a2-ab-ac+a2) - (b+c)(b2-bc+c2)
=(b+c)(a2+ab+ac+bc)
b) x3+y3+z3-3xyz
= (x+y)3-3xy(x+y) +z3-3xyz
= (x+y+z)(x2+y2+2xy-xz-yz+z2) - 3xy(x+y+z)
=(x+y+z)( x2+y2+z2-xy-yz-xz)
ta có :
\(a^3+c^3=\left(a+c\right)^3-3ac\left(a+c\right)\)
nên \(a^3+c^3-b^3+3abc=\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)\)
\(=\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2-3ac\right]=\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)\)
b. tương tự ta có :
\(a^3-b^3-c^3-3abc=a^3-\left(b+c\right)^3+3bc\left(b+c-a\right)\)
\(=\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2-3bc\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+3\left(x-z\right)\left(z-y\right)\left(x-y\right)+\left(z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=3\left(x-z\right)\left(z-y\right)\left(x-y\right)\)
#)Giải :
a) \(x+y+z=0\Leftrightarrow x+y=-z\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\Leftrightarrow x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)\) hay 3xyz (đpcm)
b) \(x=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(\Leftrightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) (Áp dụng hằng đẳng thức)
\(\Leftrightarrow x=\left[\left(b-c\right)^3+\left(c-a\right)^3\right]+\left(a-b\right)^3\)
\(=\left[\left(b-a\right)^3+\left(c-a\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[\left(b-c\right)+\left(c-a\right)\right]+\left(a-b\right)^3\)
\(=\left(b-a\right)^3-3\left(b-c\right)\left(c-a\right)\left(b-a\right)+\left(a-b\right)^3\)
\(=\left[-\left(a-b\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[-\left(a-b\right)\right]+\left(a-b\right)^3\)
\(=-\left(a-b\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)+\left(a-b\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)