Prove that: x - 1 + y - 1 - 1 = x y x + y

We need to prove, x - 1 + y - 1 - 1 = x y x + y . Using a - m = 1 a m , where a is real number and m is a positive integer. L.H.S = x - 1 + y - 1 - 1 = 1 x - 1 + y - 1 = 1 1 x + 1 y = 1 x + y x y = x y x + y = R.H.S Hence, proved.

Prove that: x - 1 + y - 1 - 1 = x y x + y

We need to prove, x - 1 + y - 1 - 1 = x y x + y . Using a - m = 1 a m , where a is real number and m is a positive integer. L.H.S = x - 1 + y - 1 - 1 = 1 x - 1 + y - 1 = 1 1 x + 1 y = 1 x + y x y = x y x + y = R.H.S Hence, proved.

Prove that: a + b + c a - 1 b - 1 + b - 1 c - 1 + c - 1 a - 1 = a b c

Here we need to prove: a + b + c a - 1 b - 1 + b - 1 c - 1 + c - 1 a - 1 = a b c Simplify L.H.S by using law: - a - m = 1 a m L.H.S = a + b + c a - 1 b - 1 + b - 1 c - 1 + c - 1 a - 1 = a + b + c 1 a b + 1 b c + 1 c a = a + b + c a + b + c a b c = 1 1 a b c = a b c = R.H.S Hence, proved

Prove that: 1 1 + x b - a + x c - a + 1 1 + x a - b + x c - b + 1 1 + x b - c + x a - c = 1

Here we need to prove: 1 1 + x b - a + x c - a + 1 1 + x a - b + x c - b + 1 1 + x b - c + x a - c = 1 Simplify L.H.S using identity, a m - n = a m a n . L.H.S = 1 1 + x b x a + x c x a + 1 1 + x a x b + x c x b + 1 1 + x b x c + x a x c = 1 x a + x b + x c x a + 1 x b + x a + x c x b + 1 x c + x b + x a x c = x a x a + x b + x c + x b x b + x a + x c + x c x c + x b + x a = x a + x b + x c x a + x b + x c = 1 = RHS Hence, Proved.

Prove that: x a x b a b . x b x c b c . x c x a c a = 1

Here we have to prove x a x b a b . x b x c b c . x c x a c a = 1 We can simplify LHS by using identity- x p q = ( x p ) 1 q LHS = x a x b a b . x b x c b c . x c x a c a = x a x b 1 a b . x b x c 1 b c . x c x a 1 c a = x a 1 a b x b 1 a b . x b 1 b c x c 1 b c . x c 1 c a x a 1 c a = x 1 b x 1 a . x 1 c x 1 b . x 1 a x 1 c = 1 = RHS Hence, proved.

If 3 x = 5 y = 45 z , prove that x = 2 y z y - z .

Let, 3 x = 5 y = 45 z = k We have a m = b ⇒ a = b 1 m , then 3 = k 1 x , 5 = k 1 y , 45 = k 1 z Now, 45 = 5 × 3 × 3 ⇒ k 1 z = k 1 y . k 1 x . k 1 x ⇒ k 1 z = k 1 y . k 2 x ⇒ k 1 z k 1 y = k 2 x ⇒ k 1 z - 1 y = k 2 x ⇒ k y - z y z = k 2 x ⇒ y - z y z = 2 x ⇒ x = 2 y z y - z Hence, proved.

If x = a b p - 1 , y = a b q - 1 and z = a b r - 1 , prove that x q - r × y r - p . z p - q = 1 .

Given: x = a b p - 1 , y = a b q - 1 and z = a b r - 1 Use the identity: a m . a n = a m + n to simplify LHS. LHS = x q - r × y r - p . z p - q ∵ a m - n = a m a n ⇒ x q x r × y r y p . z p z q = a q b p q - q a r b p r - r . a r b q r - r a p b p q - p . a p b p r - p a q b q r - q ⇒ a q b p q - q a r b p r - r . a r b q r - r a p b p q - p . a p b p r - p a q b q r - q = a p + q + r . b p q + q r + p r - ( p + q + r ) a p + q + r . b p q + q r + p r - ( p + q + r ) ⇒ 1 = RHS Hence, proved.