solve it guys it's urgent
Answers
Answer:
Given, x+y+z=8,xy+yz+zx=20
Now, x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
Again, (x+y+z)
2
=x
2
+y
2
+z
2
+2xy+2yz+2zx
x
2
+y
2
+z
2
=(x+y+z)
2
−2(xy+yz+zx)
x
2
+y
2
+z
2
=(8)
2
−2(20)=24
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=(8)[24−(20)]
x
3
+y
3
+z
3
−3xyz=(8)[24−(20)]=8(4)=32
hope it helps
sorry for bad typing
Squaring, x + y + z = 8 both sides, we get (x + y + z)2 = (8)2 x2 + y2 + z2 + 2(xy + yz + zx) = 64 x2 + y2 + z2 + 2 x 20 = 64 x2 + y2 + z2 + 40 = 64 x2 + y2 + z2 = 24 Now, x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] = 8(24 – 20) = 8 x 4 = 32 ⇒ x3 + y3 + z3 – 3xyz = 32Read more on Sarthaks.com - https://www.sarthaks.com/604661/if-x-y-z-8-and-xy-yz-zx-20-find-the-value-of-x-3-y-3-z-3-3xyz