If a b = c d = e f , prove that 2 a 4 b 2 + 3 a 2 e 2 - 5 e 4 f 2 b 6 + 3 b 2 f 2 - 5 f 5 = a 4 b 4 .

Given that a b = c d = e f . Hence, a b = e f . ⇒ a f = e b ⇒ e 2 b 2 = a 2 f 2 (squaring both sides) ⇒ e 2 = a 2 f 2 b 2 . . . 1 ⇒ e 2 a 2 = a 4 f 2 b 2 ; Also, from the equation 1 we can say e 4 f = a 4 f 5 b 4 . Putting these values in the given expression, we have 2 a 4 b 2 + 3 · a 4 f 2 b 2 - 5 a 4 f 5 b 4 2 b 6 + 3 b 2 f 2 - 5 f 5 = 2 b 6 + 3 b 2 f 2 - 5 f 5 a 4 b 4 2 b 6 + 3 b 2 f 2 - 5 f 5 = a 4 b 4 .

If a b = b c = c d , prove that a d is equal to a 5 + b 2 c 2 + a 3 c 2 b 4 c + d 4 + b 2 c d 2 .

Given that a b = b c = c d . If a b = c d , then b c = a d . Hence, b 2 c 2 = a 2 d 2 . . . i If a b = b c , then b 2 = a c . We can write b 4 c 2 = b 2 c 2 × b 2 = a 2 d 2 × a c ⇒ b 4 c = a 3 d 2 . . . ii Now putting these values of ( i ) and ( ii ) in a 5 + b 2 c 2 + a 3 c 2 b 4 c + d 4 + b 2 c d 2 = a 5 + a 2 d 2 + a 3 c 2 a 3 d 2 + d 4 + b 2 c d 2 = a 2 a 3 + d 2 + a c 2 d 2 a 3 + d 2 + b 2 c = a d a 3 + d 2 + a c 2 a 3 + d 2 + a c 2 (since b 2 = a c and b 2 c = a c 2 ) = a d

If x q + r - p = y r + p - q = z p + q - r ; show that ( q - r ) x + ( r - p ) y + ( p - q ) z = 0 .

Let x q + r - p = y r + p - q = z p + q - r = k (say) ∴ x = k ( q + r - p ) ; y = k ( r + p - q ) ; z = k ( p + q - r ) Now ( q - r ) x + ( r - p ) y + ( p - q ) z = ( q - r ) · k ( q + r - p ) + ( r - p ) · k ( r + p - q ) + ( p - q ) · k ( p + q - r ) = k [ ( q - r ) ( q + r - p ) + ( r - p ) ( r + p - q ) + ( p - q ) ( p + q - r ) ] = q 2 - r 2 - p q + p r + r 2 - p 2 - q r + p q + p 2 - q 2 - p r + q r = k [ 0 ] = 0

If y x - z = y + x z = x y , find the ratio of x : y : z .

We know that if l n = a b = c d , then l n = a b = c d = l + a + c n + b + d . So, we can say that y x - z = y + x z = x y = y + y + x + x x - z + z + y = 2 ( x + y ) x + y = 2 ∴ y x - z = y + x z = x y = 2 Taking first two terms of the ratios we get, y x - z = y + x z . ⇒ y z = x y + x 2 - z y - z x ⇒ 2 y z = x y + x 2 - z x ⇒ 2 y z + z x = x ( y + x ) ⇒ 4 y z = 2 y ( x + y ) or x = 2 y from last two or x + y - 2 z = 0 . . . ( i ) Taking first and last, we have y = 2 x - 2 z or 2 x - y - 2 z = 0 . . . ( ii ) From ( i ) and ( ii ) by cross multiplication, x - 4 = y - 2 = z - 3 or x 4 = y 2 = z 3

If y + z p b + q c = z + x p c + q a = x + y p a + q b , then show that 2 ( x + y + z ) a + b + c = ( b + c ) x + ( c + a ) y + ( a + b ) z b x + c a + a b .

y + z p b + q c = z + x p c + q a = x + y a p + q b y + z p b + q c = z + x p c + q a = x + y a p + q b = 2 ( x + y + z ) p ( a + b + c ) + q ( a + b + c ) = 2 ( x + y + z ) ( p + q ) ( a + b + c ) Now multiplying the first fraction by a in the numerator and denominator and likewise second by b , third by c and adding as above, we have each of the above ratio equal to = a y + a z + b z + b x + c x + c y a b p + a c q + b c p + a b q + c a p + b c q = x ( b + c ) + y ( c + a ) + z ( a + b ) p ( a b + b c + a c ) + q ( a b + b c + a c ) = x ( b + c ) + y ( c + a ) + z ( a + b ) ( p + q ) ( a b + b c + a c ) ∴ 2 ( x + y + z ) ( p + q ) ( a + b + c ) = x ( b + c ) + y ( c + a ) + z ( a + b ) ( p + q ) ( a b + b c + a c ) Hence, 2 ( x + y + z ) a + b + c = x ( b + c ) + y ( c + a ) + z ( a + b ) a b + b c + a c .