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Ta có
\(\left(x+y\right)^3-\left(x-y\right)^3=x^3+3x^2y+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
\(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)=2y\left(3x^2+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= (x +y)3 - ( x3+y3) = (x+y)(( x+y)2 - (x2 -xy +y2)) =3xy(x+y)
\(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]\)
\(=3xy\left(x+y\right)\)
~ rất vui vì giúp đc bn ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+y\right)^3-1-3\left(x+1\right)\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3\left(x+1\right)\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-\left(3x+3\right)\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3x-3\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2-2x+y-2\right)\)
bạn kiểm tra lại nhé :)
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Hướng dẫn
Đặt là x,y,z
Chứng minh được là \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
=(x+y+z-x)[(x+y+z)2+x(x+y+z)+x2)-(y+z)(y2-yz+z2)
=(y+z)(x2+y2+z2 +x2+xy +xz+x2 +2xy +2yz +2xz) -(y+z)(y2-yz+z2)
=(y+z)(3x2+3xy+3yz+3xz)
=3(y+z)(x+y)(x+z)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]\)\(-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+xy+xz+x^2\right)\)\(-\left(y+z\right)\left(y^2-z+z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3yz+3xz\right)\)
\(=3\left(y+z\right)\left(x+y\right)\left(y+z\right)\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+y+z\right)^3=x^3+y^3+z^3+\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(\left(x+y+z\right)^3-x^3+y^3+z^3\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(x^3-y-x-y^3=\left(x^3-x\right)-\left(y+y^3\right)\)
\(=x\left(x^2-1\right)-y\left(1+y^2\right)\)
\(=x\left(x-1\right)\left(x+1\right)-y\left(1+y^2\right)\)