Question

Chứng minh đẳng thức:

( x + y + z )³ = x³ + y³ + z³ + 3( x + y )( x+ z )( y + z)

Answer 1

\(VT=\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3\)

\(=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)

\(=x^3+y^3+3xy\left(x+y\right)+z^3+3\left(x+y\right)z\left(x+y+z\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

\(=VP\left(đpcm\right)\)

Answer 2

\(\left(x+y+z\right)^3=x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3z^2x+3x^2z+3z^2x+6xyz\)

=\(x^3+y^3+z^3+3\left(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+2xyz\right)\)

=\(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)(đpcm)

Answer 3

\(VT=\left(x+y+z\right)^{^3}=\left(x+y+z\right)\left(x+y+z\right)\left(x+y+z\right)\)

Nhân đa thức với đa thức lại với nhau :

\(\Rightarrow=VP\left(đpcm\right)\)

Answer 4

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