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Question

Verify that $${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 2 }-3x{ y }^{ 2 }=\dfrac { 1 }{ 2 } \left( x+y+z \right) \left[ { \left( zx-y \right)  }^{ 2 }+{ \left( y-z \right)  }^{ 2 }+{ \left( z-x \right)  }^{ 2 } \right]$$


Solution

$${x^3} + {y^3} + {z^2} - 3xyz$$
$$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
$$ = \dfrac{1}{2}(x + y + z)(2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx)$$
$$ = \dfrac{1}{2}(x + y + z)({x^2} + {y^2} - 2xy + {y^2} + {z^2} - 2yz + {x^2} + {z^2} - 2zx)$$
$$ = \dfrac{1}{2}(x + y + z)[{(x - y)^2} + {(y - z)^2} + {(z - x)^2}]$$

Mathematics

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