Solve: x - 3 x + 3 - x + 3 x - 3 + 6 6 7 = 0

Given x - 3 x + 3 - x + 3 x - 3 + 6 6 7 = 0 We have to find the solution of given quadratic equation. ∵ x - 3 x + 3 - x + 3 x - 3 + 6 6 7 = 0 Putting x - 3 x + 3 = y in the given equation We get y - 1 y + 6 6 7 = 0 ⇒ y 2 - 1 y = - 48 7 ⇒ 7 y 2 - 7 = - 48 y ⇒ 7 y 2 + 48 y - 7 = 0 ⇒ 7 y 2 + 49 y - y - 7 = 0 ⇒ 7 y ( y + 7 ) - 1 ( y + 7 ) = 0 ⇒ ( y + 7 ) ( 7 y - 1 ) = 0 Now if y + 7 = 0 then y = - 7 and if 7 y - 1 = 0 then y = 1 7 ∵ y = - 7 ∴ x - 3 x + 3 = - 7 or, x - 3 = - 7 x - 21 or, 8 x = - 18 ∴ x = - 18 8 = - 9 4 . ∵ y = 1 7 ∴ x - 3 x + 3 = 1 7 or, 7 x - 21 = x + 3 or, 6 x = 24 ∴ x = 24 6 = 4 ∴ The required solution is x = - 9 4 and x = 4 .

If x = 7 + 3 7 - 3 and x y = 1 , prove that x 2 + x y + y 2 x 2 - x y + y 2 = 12 11

Given that x = 7 + 3 7 - 3 Also, x y = 1 ⇒ y = 1 x ∴ y = 7 - 3 7 + 3 x + y = 7 + 3 7 - 3 + 7 - 3 7 + 3 ⇒ x + y = 7 + 3 + 2 21 + 7 + 3 - 2 21 7 - 3 ⇒ x + y = 20 4 ⇒ x + y = 5 Now, x - y = 7 + 3 7 - 3 - 7 - 3 7 + 3 ⇒ x - y = 7 + 3 + 2 21 - 7 - 3 + 2 21 7 - 3 ⇒ x - y = 4 21 4 ⇒ x - y = 21 Now, we have x 2 + x y + y 2 x 2 - x y + y 2 = x + y 2 - x y x - y 2 + x y = 5 2 - 1 21 2 + 1 = 25 - 1 21 + 1 = 24 22 = 12 11 Hence, proved

If a ∝ b , b ∝ c , prove that a 3 b 3 + b 3 c 3 + c 3 a 3 ∝ a b c a 3 + b 3 + c 3

Given, a ∝ b ⇒ a = b k [ k is non-zero variation constant] b ∝ c ⇒ b = c m [ m is non-zero variation constant] So, a = c m k Now, we have a 3 b 3 + b 3 c 3 + c 3 a 3 ∝ a b c a 3 + b 3 + c 3 ⇒ a 3 b 3 + b 3 c 3 + c 3 a 3 a b c a 3 + b 3 + c 3 = c m k 3 c m 3 + c m 3 c 3 + c 3 c m k 3 c m k × c m × c c m k 3 + c m 3 + c 3 ⇒ a 3 b 3 + b 3 c 3 + c 3 a 3 a b c a 3 + b 3 + c 3 = c 6 m k 3 m 3 + m 3 + m k 3 c 6 m k × m × m k 3 + m 3 + 1 ⇒ a 3 b 3 + b 3 c 3 + c 3 a 3 a b c a 3 + b 3 + c 3 = m k 3 m 3 + m 3 + m k 3 m k × m × m k 3 + m 3 + 1 ⇒ a 3 b 3 + b 3 c 3 + c 3 a 3 a b c a 3 + b 3 + c 3 = Non-zero variation constant Hence, a 3 b 3 + b 3 c 3 + c 3 a 3 ∝ a b c a 3 + b 3 + c 3

If b z + c y a = z x + a z b = a y + b x c , then show that x a b 2 + c 2 - a 2 = y b c 2 + a 2 - b 2 = z c a 2 + b 2 - c 2

Given, b z + c y a = z x + a z b = a y + b x c By interchanging the terms, we get, a b z + c y = b z x + a z = c a y + b x Let's multiply the numerator and denominator by the numerator terms. a 2 a b z + a c y = b 2 b z x + a b z = c 2 a c y + b c x By using the addendo process we have, ⇒ b 2 + c 2 - a 2 b c x + a b z + a c y + b c z - a b z - a c y ⇒ b 2 + c 2 - a 2 2 b c x Similarly, ⇒ a 2 + b 2 - c 2 2 a b z and c 2 + a 2 - b 2 2 a c y Now, let's equate these to each other we get, ⇒ b 2 + c 2 - a 2 2 b c x = a 2 + b 2 - c 2 2 a b z = c 2 + a 2 - b 2 2 a c y ⇒ b 2 + c 2 - a 2 b c x = a 2 + b 2 - c 2 a b z = c 2 + a 2 - b 2 a c y Multiply by a b c we get, ⇒ a b 2 + c 2 - a 2 x = b a 2 + b 2 - c 2 z = c c 2 + a 2 - b 2 y By interchanging the terms, we get, x a b 2 + c 2 - a 2 = y b c 2 + a 2 - b 2 = z c a 2 + b 2 - c 2 .