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SSC CHSL 5-March-2018 Shift 1 Paper
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© examsnet.com
Question : 3
Total: 100
Determine the value of
(
1
r
+
1
s
)
when
r
3
+
s
3
=
0
and
r
+
s
=
6
0
0.5
1
6
Validate
Solution:
Given :
r
3
+
s
3
=
0
and
r
+
s
=
6
---------(i)
∵
x
3
+
y
3
=
(
x
+
y
)
(
x
2
+
y
2
−
x
y
)
⇒
(
r
+
s
)
(
r
2
+
s
2
−
r
s
)
=
0
⇒
6
(
r
2
+
s
2
−
r
s
)
=
0
⇒
r
2
+
s
2
−
r
s
=
0
⇒
r
2
+
s
2
=
r
s
----------(ii)
Also, squaringequation (i),
⇒
(
r
+
s
)
2
=
(
6
)
2
⇒
r
2
+
s
2
+
2
r
s
=
36
Substituting value from equation (ii), ⇒ rs + 2rs = 3rs = 36
⇒
r
s
=
36
3
=
12
------------(iii)
∴
(
1
r
+
1
s
)
=
r
+
s
r
s
Dividing equation (i) by (iii),
=
6
12
=
0.5
© examsnet.com
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