Question

prove that x cube + y cube + Z cube minus 3 x y z is equal to X + y +Z X square + Y square + Z square minus xy -yz minus z x​

Answer 1

we have to prove that ,

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

proof : LHS = x³ + y³ + z³ - 3xyz

= (x³ + y³) + z³ - 3xyz

[from algebraic identity, a³ + b³ = (a + b)³ - 3ab(a + b) ]

= (x + y)³ - 3xy(x + y) + z³ - 3xyz

= {(x + y)³ + z³} - 3xy(x + y) - 3xyz

= {(x + y + z)³ - 3z(x + y)(x + y + z)} - 3xy(x + y) - 3xy

= (x + y + z)³ - 3z(x + y)(x + y + z)-3xy(x + y) - 3xyz

= (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 ]

[ from algebraic identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ]

= (x + y + z)[ x² + y² + z² + 2xy + 2yz + 2zx - 3zx - 3zy - 3xy ]

= (x + y + z) [ x² + y² + z² - xy - yz - zx ] = RHS