(x+y)²/ xy + (y+z)1/yz (z+x)1/xz if x+y+z = 0
Answers
Answered by
0
Step-by-step explanation:
Given
x+y+z=1,xy+yz+zx=−1 and xyz=−1
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
−3(−1)=1(x
2
+y
2
+z
2
)−(−1)
x
3
+y
3
+z
3
+3=x
2
+y
2
+z
2
+1
x
3
+y
3
+z
3
+2=x
2
+y
2
+z
2
(x+y+z)
2
=x
2
+y
2
+z
2
+2xy+2yz+2xz
1
2
=x
2
+y
2
+z
2
+2(−1)
x
2
+y
2
+z
2
=1+2
x
2
+y
2
+z
2
=3
x
3
+y
3
+z
3
+2=3
x
3
+y
3
+z
3
=1
Answered by
0
Answer:
Given
x+y+z=1,xy+yz+zx=−1 and xyz=−1
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
−3(−1)=1(x
2
+y
2
+z
2
)−(−1)
x
3
+y
3
+z
3
+3=x
2
+y
2
+z
2
+1
x
3
+y
3
+z
3
+2=x
2
+y
2
+z
2
(x+y+z)
2
=x
2
+y
2
+z
2
+2xy+2yz+2xz
1
2
=x
2
+y
2
+z
2
+2(−1)
x
2
+y
2
+z
2
=1+2
x
2
+y
2
+z
2
=3
x
3
+y
3
+z
3
+2=3
x
3
+y
3
+z
3
=1
Step-by-step explanation:
hope this helps you
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