a+b+c=0 then proved that xyz=1
Answers
Answer:
From the following, we need to find the values of x,y, and z:
a x =b , b y =c , c z =a
From ax = b, we get x =b/a -------------------------- (i)
From by = c, we get y =c/b ---------------------------(ii)
From cz = a, we get z = c/a ---------------------------(iii)
Multiplying the left hand sides and right hand sides of (i), (ii) and (iii) we get
xyz=(b/a)*(c/b)*(c/a)
When we simplify, we find that xyz=1. Hence proved.
Step-by-step explanation:
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Answer:
Given:- xyz=1
To prove:- (1+x+y
−1
)
−1
+(1+y+z
−1
)
−1
+(1+z+x
−1
)
−1
=1
Proof:-
Taking L.H.S.-
(1+x+y
−1
)
−1
+(1+y+z
−1
)
−1
+(1+z+x
−1
)
−1
=(1+x+xz)
−1
+(1+y+xy)
−1
+(1+z+yz)
−1
=(1+x+xz)
−1
+(xyz+y+xy)
−1
+(1+z+yz)
−1
=(1+x+xz)
−1
+y
−1
(1+x+xz)
−1
+(1+z+yz)
−1
=(1+x+xz)
−1
(1+y
−1
)+(1+z+yz)
−1
=(xyz+x+xz)
−1
(1+y
−1
)+(1+z+yz)
−1
=x
−1
(1+z+yz)
−1
(1+y
−1
)+(1+z+yz)
−1
=(1+z+yz)
−1
(x
−1
+(xy)
−1
)+(1+z+yz)
−1
=(1+z+yz)
−1
(yz+z+1)
=
1+z+yz
1+z+yz
=1
= R.H.S.