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Byju's Answer
Standard VIII
Mathematics
Algebraic Identity (a-b)^3
r1 - r r2 ...
Question
(
r
1
−
r
)
(
r
2
−
r
)
(
r
3
−
r
)
=
4
r
2
R
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Solution
(
r
1
−
r
)
(
r
2
−
r
)
(
r
3
−
r
)
=
(
r
1
r
2
−
r
r
1
−
r
r
2
+
r
2
)
(
r
3
−
r
)
=
r
1
r
2
r
3
−
r
r
1
r
3
−
r
r
2
r
3
+
r
2
r
3
−
r
r
1
r
2
+
r
2
r
1
+
r
2
r
2
−
r
3
=
s
2
r
−
r
(
r
1
r
3
+
r
2
r
3
+
r
1
r
3
)
+
r
2
(
r
1
+
r
2
+
r
3
)
−
r
3
=
s
2
r
−
r
(
s
2
)
+
r
2
(
4
R
+
r
)
−
r
2
=
s
2
r
−
s
2
r
+
4
R
r
2
+
r
3
−
r
3
=
4
r
2
R
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Similar questions
Q.
Prove that
(
r
1
−
r
)
(
r
2
−
r
)
(
r
3
−
r
)
=
4
R
r
2
where In
△
A
B
C
,
r
and
R
are inradius and circumradius and
r
1
,
r
2
,
r
3
are exradius respectively.
Also,
a
,
b
,
c
are the corresponding sides.
Q.
(
r
1
−
r
)
(
r
2
+
r
3
)
=
a
2
Q.
In a
△
A
B
C
,
(
r
1
+
r
3
)
√
r
r
2
r
1
r
3
is equivalent to
Q.
In a
△
A
B
C
,
r
r
2
+
r
1
r
3
=
Q.
Two resistors of resistance
R
1
and
R
2
having
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1
>
R
2
are connected in parallel. For equivalent resistance R, the correct statement is:
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