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UPSC CDS 1 2018 Math Paper
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© examsnet.com
Question : 10
Total: 100
If
x
+
y
+
z
=
0
,
then what is
(
y
+
z
−
x
)
2
+
(
z
+
x
−
y
)
3
+
(
x
+
y
−
z
)
3
equal to ?
[2018 CDS-I]
(
x
+
y
+
z
)
3
3
(
x
+
y
)
(
y
+
z
)
(
z
+
x
)
24
x
y
z
−
24
x
y
z
Validate
Solution:
It is known that
If
x
+
y
+
z
=
0
,then
x
3
+
y
3
+
z
3
=
3
x
y
z
now
(
y
+
z
−
x
)
3
+
(
z
+
x
−
y
)
3
+
(
x
+
y
−
z
)
3
=
(
−
x
−
x
)
3
+
(
−
y
−
y
)
3
+
(
−
z
−
z
)
3
=
(
−
2
x
)
3
+
(
−
2
y
)
3
+
(
−
2
z
)
3
=
−
8
(
x
3
+
y
3
+
z
3
)
=
−
8
(
3
x
y
z
)
=
−
24
x
y
z
Alternative Method:
Take x = 1, y = 1, z = –2. Then
y
+
z
−
x
)
3
+
(
z
+
x
−
y
)
3
+
(
x
+
y
z
3
=
(
1
−
2
−
3
)
3
+
(
−
2
+
1
−
1
)
3
=
(
1
+
1
+
2
z
)
3
=
(
−
8
)
+
(
−
8
)
+
64
=
48
Also
−
24
x
x
y
z
=
−
24
×
1
×
1
×
(
−
2
)
=
48
Therefore
(
y
+
z
−
x
)
3
+
(
z
+
x
−
y
)
3
+
(
x
+
y
−
z
)
3
=
−
24
x
y
z
© examsnet.com
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