Question

x+y+z=1 then find value 1-3x^2-3y^2-3z^3+2x^3+2y^3+2z^3

Answer 1

HELLO DEAR,

given that:-

(x + y + z ) = 1-------------(1)



1 - 3x² - 3y² - 3z² + 2x³ + 2y³ + 2z³

=> 1 - 3x² - 3y² - 3z² + 2[(x³ + y³ +z³) ]

=> 1 - 3x² - 3y² - 3z² + 2[(x+y+z)(x²+y²+z² - xy - yz - zx) + 3xyz]-------from (1)

=> 1 - 3x² - 3y² - 3z² + 2[(1)(x²+y²+z² - xy - yz - zx) + 3xyz]

=> 1 - 3x² - 3y² - 3z² + 2x² + 2y² + 2z² - 2xy - 2yz - 2zx + 6xyz]

=> 1 - x² - y² - z² - 2xy - 2yz - 2zx + 6xyz

=> 1 - (x² + y² +z² + 2xy + 2yz + 2zx) + 6xyz

=> 1 - (x+y+z)² + 6xyz---------from(1)

=> 1 - 1 + 6xyz

=> 6xyz



I HOPE ITS HELP YOU DEAR,
THANKS