If x=1, y=2 and z=3, then find the value of: (x2 – 22)+(x2 - y2)+(12 – z2)+(22-x²)+(z2 - y2) +(12-x2)
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Answer:
866
Given x+y+z=12
Squaring it we get
x
2
+y
2
+z
2
+2(xy+yz+zx)=144
Therefore xy+yz+zx=24
Also 1/x+1/y+1/z=36
Implies (xy+yz+zx)/xyz=36
⇒xyz=(xy+yz+zx)/36=24/36=2/3
We know (x
3
+y
3
+z
3
)=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)+3xyz
⇒x
3
+y
3
+z
3
=(12)(96−24)+2=12∗72+2=866
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