Question

Prove that x³+y³+z³-3xyz=1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]​

Answer 1

Answer:

SOLUTION

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

=1/2 (x + y + z){2x² + 2y² + 2z² -2xy - 2yz - 2zx }

= 1/2 (x + y + z){( x² -2xy + y²) + (y² - 2yz + z²) +(z² - 2zx + x²) }

= 1/2 (x + y + z){(x - y)² + (y - z)² + (z - x)²}

Hence,verified

Now,

64x³ + 125y³ - 64z³ +240xyz

= (4x)³ + (5y)³ +(-4z)³ -3(4x)(5y)(-4z)

This is just like above Identity , use that .

=1/2 (4x + 5y -4z ){(4x -5y)² + (5y+4z)² +(4x + 4z)² }

Hence, (4x + 5y -4z ) and {(4x -5y)² + (5y+4z)² +(4x + 4z)² } are the factors of given expression.