If A, B, C are in continue proportion the prove abc(a+B+C) ^3 = (ab+bc+cd)^3
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a,b,c,d are in continued proportion
⇒
b
a
=
c
b
=
d
c
=k
⇒a=bk,b=ck,c=dk,
To prove a
2
+b
2
+c
2
)(b
2
+c
2
+d
2
)=(ab+bc+cd+)
2
substituting a,b,c
⇒((bk)
2
+(ck)
2
+(dk)
2
)(b
2
+c
2
+d
2
)=(b
2
k+c
2
k+d
2
k)
2
⇒k
2
(b
2
+c
2
+d
2
)(b
2
+c
2
+d
2
)=k
2
(b
2
+c
2
+d
2
)
2
⇒(b
2
+c
2
+d
2
)
2
=(b
2
+c
2
+d
2
)
2
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