If a , b , c are in continued proportion, show that: a 2 + b 2 b ( a + c ) = b ( a + c ) b 2 + c 2 .

As a , b , c are in continued proportion, So, a b = b c b 2 = a c ....(i) Now, ( a 2 + b 2 ) ( b 2 + c 2 ) = ( a 2 + a c ) ( a c + c 2 ) (Using (i)) ( a 2 + b 2 ) ( b 2 + c 2 ) = a ( a + c ) c ( a + c ) ( a 2 + b 2 ) ( b 2 + c 2 ) = a c ( a + c ) 2 ( a 2 + b 2 ) ( b 2 + c 2 ) = b 2 ( a + c ) 2 (Using (i)) Thus, ( a 2 + b 2 ) ( b 2 + c 2 ) = b ( a + c ) b ( a + c ) Hence, a 2 + b 2 b ( a + c ) = b ( a + c ) b 2 + c 2

If a b = c d , show that: a 3 c + a c 3 b 3 d + b d 3 = ( a + c ) 4 ( b + d ) 4 .

Assume that, a b = c d = x ⇒ a = b x and c = d x Now, L . H . S . = a 3 c + a c 3 b 3 d + b d 3 = a c ( a 2 + c 2 ) b d ( b 2 + d 2 ) = ( b x × d x ) ( b 2 x 2 + d 2 x 2 ) b d ( b 2 + d 2 ) = x 2 × x 2 ( b 2 + d 2 ) ( b 2 + d 2 ) = x 4 ∴ R . H . S . = ( a + c ) 4 ( b + d ) 4 = ( b x + d x ) 4 ( b + d ) 4 = x ( b + d ) ( b + d ) 4 = x 4 Thus, LHS = RHS Hence, a 3 c + a c 3 b 3 d + b d 3 = ( a + c ) 4 ( b + d ) 4