Question

Prove that
(x+y)^3+(y+2)^3+(z+x)^3-3{(x+y)(y+z)(z+x)}=2x^3+y^3+z^3 -3xyz​

Answer 1

Step-by-step explanation:

Heya mate,Here is ur answer

(x+y)^3 +(y+z)^3 +(z+x)^3 -3((x+y) (y+z)(z+x)

=x^3 +y^3 +3x^2y +3 xy^2 + y^3 +z^3 +3 y^2z + 3yz^2 +z^3 +x^3 +3z^2x + 3 x^2 z - 3{( xy+xz+y^2 +yz)(z+x)}

= x^3 +y^3 +3x^2y +3 xy^2 + y^3 +z^3 +3 y^2z + 3yz^2 +z^3 +x^3 +3z^2x + 3 x^2 z - 3{xyz+xz^2 + zy^2 +yz^2 +x^2 y +x^2z+xy^2 +xyz}

=x^3 +y^3 +3x^2y +3 xy^2 + y^3 +z^3 +3 y^2z + 3yz^2 +z^3 +x^3 +3z^2x + 3 x^2 z -3{2xyz + xz^2 + zy^2 + yz^2 + x^2 y +x^2z +xy^2}

=x^3 +y^3 +3x^2y +3 xy^2 + y^3 +z^3 +3 y^2z + 3yz^2 +z^3 +x^3 +3z^2x + 3 x^2 z -6xyz - 3xz^2 - 3zy^2 - 3yz^2 -3x^2y-3x^2z -3xy^2

= x^3 +x^3 +y^3 +y^3 +z^3 +z^3 +3x^2y -3x^2y +3xy^2 -3xy^2 + 3y^2z -3y^2z +3yz^2 -3yz^2 + 3z^2x-3z^2x +3x^2z -3x^2z-6xyz

= 2x^3+2y^3 +2z^3 -6xyz

=2(x^3+y^3+z^3-3xyz)

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