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07-02-2019
Mathematics

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if x+y+z=0 Prove that (x^2+y^2+z^2)=4(xy+xz+yz)^2

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15-02-2019

x + y + z = 0

(x + y + z)² = 0

x² + y² + z² + 2xy + 2xz + 2yz = 0

x² + y² + z² = -2(xy + xz + yz)

(x² + y² + z²)² = 4(xy + xz + yz)²
eh, not same

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