Question

if x+y+z=6and x²+y²+z²=38 then what is the value of x(y²+z²)+y(x²+z²)+z(y²+x²) =?​

Answer 1

Answer:

I can solve the fast one x+y+z=6

here,x=y=z

=)x+y+z=6

=)x+x+x=6

=)3x=6

=)x=6/3

=)x=2.

Answer 2

x+y+z=6

x^2+y^2+z^2=38

Now,we have a formulae (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)

Putting known values in above formulae,we get

36=38+2(xy+yz+zx) or (xy+yz+zx)=-2/2=-1

We have a formulae -

x^3+y^3+z^3–3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

Putting known values in above equation,we get

-1–3xyz=(6)(38–(-1) {Since,we have already got the value of xy+yz+zx}

-1–3xyz=234 or 3xyz=235 or xyz=3/235

Thus,value of xyz=-78.3