1. If r + 2y + 3z = 0 and
x + 4y3 + 9z3 = 18.xyz; evaluate :
(x+2y)2
(2y + 32)2 (3z + x)2
yz
+
3z+x^2/zx
Answers
Answered by
1
Answer:
x+2y+3z=0 and x
3
+4y
3
+9z
3
=18xyz
=>x+2y=−3z
also, =>2y+3z=−x
and =>3z+x=−2y
Now,
xy
(x+2y)
2
+
yz
(2y+3z)
2
+
zx
(3z+x)
2
=
xy
(−3z)
2
+
yz
(−x)
2
+
zx
(−2y)
2
=
xy
9z
2
+
zy
x
2
+
zx
4y
2
=
xyz
9z
3
+
xyz
x
3
+
xyz
4y
3
=
xyz
9z
3
+x
3
+4y
3
=
xyz
18xyz
=18
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