Question

Verify that x³ + y³ + z³ - 3xyz = 1/2(x+y+z)[(x - y)² +(y – z)² + (z - x)²]
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Answer 1

To prove:-

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

We know that:-

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

x³+y³+z³-3xyz =1/2(x+y+z) ×2(x²+y²+z²-xy-yz-zx)

x³+y³+z³-3xyz=1/2(x+y+z)(2x²+2y²+2z²-xy-yz-zx)

=1/2(x+y+z)(x²+x²+y²+y²+z²+z²-2xy-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]

Therefore:-

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

Answer 2

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