If x2+y2+z2 = 2(x+2-1), then the value of x3+y3+z?
will be
Answers
Answer:
see that ⬇️⬇️⬇️
Step-by-step explanation:
If x, y, z be positive real numbers such that x
2
+y
2
+z
2
=27, then show that x
3
+y
3
+z
2
≥81
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ANSWER
Applying Cauchy- Schwarz inequality to the two sets of numbers
x
2
3
,y
2
3
,z
2
3
;
x
2
1
,y
2
1
,z
2
1
we have
(x
2
3
.x
2
1
+y
2
3
.y
2
1
+z
2
3
.z
2
1
)
2
≤(x
3
+y
3
+z
3
)(x+y+z)
(x
2
+y
2
+z
2
)
2
≤(x
3
+y
3
+z
3
)(x+y+z)...........(1)
Applying Cauchy Schwarz inequality to the two sets numbers x, y, z ; 1, 1, 1
(x+y+z)
2
≤(x
2
+y
2
+z
2
)(1+1+1).........(2)
Squaring (1)
(x
2
+y
2
+z
2
)
4
≤(x
3
+y
2
+z
3
)
2
(x+y+z)
2
Combining (3) and (2)
(27)
4
≤(x
3
+y
2
+z
3
)
2
(x+y+z)
2
.3
(27)
4
≤(x
3
+y
2
+z
3
)
2
.27.3
3
8
≤(x
3
+y
2
+z
3
)
2
⇒(x
3
+y
2
+z
3
)
2
≥(3
4
)
2
x
3
+y
3
+z
3
≥81.
Answer:
when ,
x²+y²+z²=2(x+z-1)
value of x³+y³+z3..
Solution-
x²+y²+z² =2(x+z-1)
x²+y²+z=2x+2z-2
x²+y²+z²=2x+2z-1-1
(x²+1-2x)+y²
+(z²+1-2z)=0
(x-1)²+y²+(z-1)²=0
(x-1)²=0
x=1
y²=0
y=0
(z-1)²=0
Value substituted in question
x³+y³+z³
1³+0+1³
=2 ANSWER
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