Question

find the value of x^3+y^3+z^3-3xyz if x^2+y^2+z^2=83 and x+y+z=15
need appropriate answer
need a proper answer for grade 9(step by step)

Answer 1

Explanation:

Given,

$x^2 + y^2 + z^2 = 83, x+y+z=15$

$(x+y+z)^2 = 15^2$

$x^2 + y^2 + z^2 +2(xy+yz+zx)=225$

$2(xy+yz+zx)+83=225$$2(xy+yz+zx)=225-83$

$(xy + yz + zx) = \frac{142}{2}$

$(xy+yz+zx)=71$

$-(xy+yz+zx)=(-xy-yz-zx)=-71$

It is known that

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

$x^3+y^3+z^3-3xyz=(15)(83-71)$

∴ $x^3+y^3+z^3-3xyz=(15)(12)=180$

Answer: 180

I hope this helps. :D

Answer 2

Answer:

Destructive local tsunamis are possible near the epicenter, and significant sea level changes and damage might occur in a broader region. Note that with a magnitude 9.0 earthquake, there is a possibility of an aftershock of magnitude 7.5 or greater.

Step-by-step explanation:

hope you are understand