1. If `x+y+z=9` and `x^2+y^2+z^2=29`, then find `xy+yz+zx`

Solution:
Here `X+Y+Z=9` and `X^2+Y^2+Z^2=29`


Now, We know that
`(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)`

`:.2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)`

`:.2(xy+yz+zx)=9^2-29`

`:.2(xy+yz+zx)=81-29`

`:.2(xy+yz+zx)=52`

`:.(xy+yz+zx)=52/2`

`:.(xy+yz+zx)=26`