Question

expand determinant (ab+bc+ca)^3

Answer 1

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

Answer 2

(ab+bc+ca)^3
(a+b+c)³=[(a+b)+c]³=(a+b)³+3(a+b)²c+3(a+b)c²+c³(a+b+c)3=(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c2+c³(a+b+c)3=a³+b³+c³+3a²b+3a²c+3ab²+3b²c+3ac²+3bc²+6abc(a+b+c)³=(a³+b³+c³)+(3a²b+3a²c+3abc)+(3ab²+3b²c+3abc)+(3ac²+3bc²+3abc)−3abc