If x+y+z=9, xy+yz+zx =26; find x²+y²+z²

*First, we must expand and simplify :*

\( x + y+z = 9 \)

\( (x + y+z)^{2} = (9)^{2} \)

\( x^{2} + y^{2}+z^{2} + 2(xy + yz+zx) = 81 \)*And Than, Enter the value :*

\( x^{2} + y^{2}+z^{2} + 2(26) = 81 \)

\( x^{2} + y^{2}+z^{2} + 52 = 81 \)

\( x^{2} + y^{2}+z^{2} = 81- 52 \)

\( \boxed{x^{2} + y^{2}+z^{2} = 29} \)

*we know,**(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)**9² = x² + y² + z² + 2 X 26** 81 = x² + y² + z² + 52 **81 - 52 = x² + y² + z²**therefore, x² + y² + z² = 29*

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