If a 3 + 3 a b 2 3 a 2 b + b 3 = x 3 + 3 x y 2 3 x 2 y + y 3 , then prove that x a = y b .

Given, a 3 + 3 a b 2 3 a 2 b + b 3 = x 3 + 3 x y 2 3 x 2 y + y 3 On applying componendo and dividendo, we get, ⇒ a 3 + 3 a b 2 + 3 a 2 b + b 3 a 3 + 3 a b 2 - 3 a 2 b - b 3 = x 3 + 3 x y 2 + 3 x 2 y + y 3 x 3 + 3 x y 2 - 3 x 2 y - y 3 ............. i We know that, a 3 + 3 a b 2 + 3 a 2 b + b 3 = a + b 3 and a 3 + 3 a b 2 - 3 a 2 b - b 3 = a - b 3 . Therefore, equation i becomes, ⇒ a + b 3 a - b 3 = x + y 3 x - y 3 Now taking cube roots on both sides, we get, ⇒ a + b a - b = x + y x - y ............ ii Again, applying componendo and dividendo, we get, ⇒ a + b + a - b a + b - a + b = x + y + x - y x + y - x + y ⇒ 2 a 2 b = 2 x 2 y ⇒ a b = x y Applying cross-multiply, we get, ⇒ b x = a y ⇒ x a = y b Hence, proved.

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

Given, a y - b x c = c x - a z b = b z - c y a Multiplying the numerator and the denominator by c , b , and a in all three ratios respectively, we get, ⇒ c a y - b x c 2 = b c x - a z b 2 = a b z - c y a 2 ⇒ c a y - c b x c 2 = b c x - b a z b 2 = a b z - a c y a 2 ................. i Clearly, each ratio = Sum of antecedents Sum of consequents = c a y - c b x + b c x - b a z + a b z - a c y c 2 + b 2 + a 2 = 0 c 2 + b 2 + a 2 = 0 ................ ii Therefore, a y - b x c = c x - a z b = b z - c y a = 0 ⇒ a y - b x = c x - a z = b z - c y = 0 Either, a y - b x = 0 or, c x - a z = 0 or, b z - c y = 0 . ⇒ a y = b x ⇒ x a = y b ........... iii ⇒ c x = a z ⇒ x a = z c ............ iv ⇒ b z = c y ⇒ y b = z c ............ v Therefore, from iii , iv and v , we get, x a = y b = z c . Hence, proved.

If a p + b q a r + b s = a p - b q a r - b s , then show that p q = r s .

Given, a p + b q a r + b s = a p - b q a r - b s Transposing the terms, we get, ⇒ a p + b q a p - b q = a r + b s a r - b s .............. i Now, applying componendo and dividendo in equation i , we get, ⇒ a p + b q + a p - b q a p + b q - a p + b q = a r + b s + a r - b s a r + b s - a r + b s ⇒ 2 a p 2 b q = 2 a r 2 b s [Cancel out the like but opposite sign terms] ⇒ a p b q = a r b s [Simplify both sides] ⇒ a p a r = b q b s [Transpose the terms] ⇒ p r = q s [Simplify both sides] ⇒ p q = r s [Transpose the terms] Hence, proved.