Question

answer this in ten min ..urgent pls​

Answer 1

Answer:

please refer to the attachment

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Answer 2

to prove,

x³ + y³ + z³ - 3xyz = ½(x + y + z)[(x - y)² + (y - z)² + (z - x)²

Step-by-step explanation:

here,

given equation

x³ + y³ + z³ - 3xyz = ½(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

and we kow that,

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

here in given equation

RHS

= ½(x + y + z)[(x - y)² + (y - z)² + (z - x)²

by solving this,

= ½(x + y + z) (x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx)

= ½(x + y + z)(2x² + 2y² + 2z² - 2xy - 2yz - 2zx)

by taking 2 as a common we get

½ × 2(x + y + z) (x² + y² + z² - 2xy - 2yz - 2zx)

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

we know that,

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

so,

x² + y² + z² - 3xyz = ½(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

proved