Question

if (x+y+z)=9 and (xy+yz+zx)=26, then find the value of x^2+y^2+z^2​

Answer 1

Step-by-step explanation:

(X+y+z)^2=x^2+y^2+z^2+2(xy+yx+yz)

81=x^2+y^2+z^2+52

81-52=x^2+y^2+z^2

x^2+y^2+z^2=29

Answer 2

Answer:

The ans. is 29

Step-by-step explanation:

According to the identity

(x+y+z)^2 =x^2+y^2+z^+2(xy+ yz+zx)

=>(9)^2=x^2+y^2+z^2+2(26)

=>81=x^2+y^2+z^2+52

=>81-52=x^2+y^2+z^2=29.