abc (a+b+c)³ = (ab+bc+ca)³
Answers
Answer:
If
{
a
,
b
,
c
}
are in a Geometric Progression with some common ratio
r
we can write the terms as
{
a
,
a
r
,
a
r
2
}
, so that:
b
=
a
r
c
=
b
r
=
a
r
2
Consider the LHS:
(
a
b
+
b
c
+
c
a
)
3
=
(
a
(
a
r
)
+
a
r
(
a
r
2
)
+
a
r
2
(
a
)
)
3
=
(
a
2
r
+
a
2
r
3
+
a
2
r
2
)
3
=
[
a
2
r
(
1
+
r
2
+
r
)
]
3
=
a
6
r
3
(
1
+
r
+
r
2
)
3
Now, consider the RHS:
a
b
c
(
a
+
b
+
c
)
3
=
a
(
a
r
)
(
a
r
2
)
(
a
+
a
r
+
a
r
2
)
3
=
a
3
r
3
[
a
(
1
+
r
+
r
2
)
]
3
=
a
3
r
3
(
a
3
)
(
1
+
r
+
r
2
)
3
=
a
6
r
3
(
1
+
r
+
r
2
)
3
And these two expression are both equal to:
a
6
r
3
(
1
+
r
+
r
2
)
3
Hence,
(
a
b
+
b
c
+
c
a
)
3
=
a
b
c
(
a
+
b
+
c
)
3
QED
abc(a+b+c)³=(ab+bc+can)³ if a,b,c are a continued proportion
a, b, c are a continued proportion
=>b=ak
&c=bk=(ak)k =ak²
LHS
=abc(a+b+c)³
=a(ak)(ak²)(a+ak+ak²)³
=a³k³a³(1+k+k²)³
=a^6k3(1+k+k²)³
RHS
=(ab+bc+can)³
=ask+akak²+ak²a)³
=(a²k+a²k3+a²k)³
=(a²k(1+k²+k))³
=(a²k)³(1+k+k²)³
=a^6k³(1+k+k²)³
LHS=RHS
QED