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a,Ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)
b, Từ:
x + y + z = 0
=> x + y = -z
<=> (x + y)^3 = (-z)^3
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2
<=> x^3 + y^3 + z^3 = -3xy(x+y)
<=> x^3 + y^3 + z^3 = -3xy(-z)
<=> x^3 + y^3 + z^3 = 3xyz
bài này 1h rùi,chắc chờ tui ngủ dậy làm;
= (x+y)3 - (x+y) + xy(x+y) =
= (x+y)((x+y)2 -1 +xy)) = (x+y)(x2 +3xy +y2 -1)
+ \(4x\left(y-1\right)-\left(y-1\right)\) \(=\left(y-1\right)\left(4x-1\right)\)
+ \(2xy-x^2-y^2+16\) \(=-\left(x^2-2xy+y^2-16\right)\) \(=-\left[\left(x-y\right)^2-4^2\right]\)\(=-\left(x-y-4\right)\left(x-y+4\right)\)
+ \(\left(x+y+z\right)^3-x^3-y^3-z^3\) \(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)\(=x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)\(=3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\) \(=3\left(x+y\right)\left(xy+xz+zy+z^2\right)\)\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
+)4x(y-1)-(y-1)
=(y-1)(4x-1)
+)2xy-x²-y²+16
=-(x²-2xy+y²)+16
=-(x-y)²+16
=-(x-y-4)(x-y+4)
Đặt \(x-y=c,y-z=a,z-x=b\) thì \(a+b+c=0\Rightarrow c=-a+b\) , ta có:
\(P=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
\(P=\left(a+b\right)\left(a^2-ab+b^2\right)+c^3\)
\(P=-c\left(a^2-ab+b^2\right)+c\left(a+b\right)^2\)
\(P=c\left(-a^2-ab+b^2+a^2+2ab+b^2\right)\)
\(P=3abc=3\left(x-y\right)\left(y-z\right)\left(z-x\right).\)