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


RELATED: