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\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
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).
\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
cộng ((x+y)^3 + z^3) vào 1 nhóm, -3xy(x+y)-3xyz vào 1 nhóm dc
\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3yz\left(x+y+z\right)\)xuất hiện nhân tử chung x+y+z
\(\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2-3xy\right)\)
Kết quả: \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a) \(x^4+5x^3+10x-4\)
\(=\left(x^4+2x^2\right)+\left(5x^3+10x\right)-\left(2x^2+4\right)\)
\(=x^2\left(x^2+2\right)+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-2\right)\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\right]\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2\right]\)
\(=\left(x^2+2\right)\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x^2+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
b) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)
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).
\(x^3+y^3+z^3+3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3+3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y+z\right)+z^3\)
\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(xy+yz+xz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3xz\right]\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
#)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)
\(\Leftrightarrow x=\left[\left(b-c\right)^3+\left(c-a\right)^3\right]+\left(a-b\right)^3\)
\(=\left[\left(b-a\right)^3+\left(c-a\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[\left(b-c\right)+\left(c-a\right)\right]+\left(a-b\right)^3\)
\(=\left(b-a\right)^3-3\left(b-c\right)\left(c-a\right)\left(b-a\right)+\left(a-b\right)^3\)
\(=\left[-\left(a-b\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[-\left(a-b\right)\right]+\left(a-b\right)^3\)
\(=-\left(a-b\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)+\left(a-b\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)