If x + y + z = 0 , prove that x + y 2 x y + y + z 2 y z + z + x 2 z x = 3

Given: x + y + z = 0 We know that, a 3 + b 3 + c 3 – 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 – a b – b c – c a ) ⇒ x 3 + y 3 + z 3 – 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 – x y – x z – z x ) ⇒ x 3 + y 3 + z 3 – 3 x y z = 0 ⇒ x 3 + y 3 + z 3 = 3 x y z We need to prove that x + y 2 x y + y + z 2 y z + z + x 2 z x = 3 LHS = x + y 2 x y + y + z 2 y z + z + x 2 z x = z x 2 + 2 x y + y 2 x y z + x y 2 + 2 y z + z 2 x y z + y z 2 + 2 z x + x 2 x y z = z x 2 + 2 x y z + y 2 z + x y 2 + 2 x y z + z 2 x + y z 2 + 2 x y z + x 2 y x y z = z x 2 + x 2 y + x y 2 + y 2 z + y z 2 + z 2 x + 6 x y z x y z = x 2 z + y + y 2 x + z + z 2 y + x + 6 x y z x y z Since, x + y + z = 0 ⇒ x + y = - z , y + z = - x and z + x = - y = x 2 - x + y 2 - y + z 2 - z + 6 x y z x y z = - x 3 - y 3 - z 3 + 6 x y z x y z = - 3 x y z + 6 x y z x y z ∵ x 3 + y 3 + z 3 = 3 x y z = 3 x y z x y z = 3 = RHS Hence, proved.

If 9 x 2 + 1 4 x 2 = 13 , find the value of 3 x + 1 2 x

We need to find the value of 3 x + 1 2 x . Given that, 9 x 2 + 1 4 x 2 = 13 Add 2 3 x 1 2 x on both sides of the equation. ⇒ 9 x 2 + 1 4 x 2 + 2 3 x 1 2 x = 13 + 2 3 x 1 2 x It is in the form of a 2 + b 2 + 2 a b where a = 3 x and b = 1 2 x ⇒ 3 x 2 + 2 3 x 1 2 x + 1 2 x 2 = 13 + 3 Apply a 2 + 2 a b + b 2 = a + b 2 ⇒ 3 x + 1 2 x 2 = 16 ⇒ 3 x + 1 2 x = ± 4

If 9 x 2 + 1 4 x 2 = 13 , find the value of 27 x 3 + 1 8 x 3 .

We need to find the value of 27 x 3 + 1 8 x 3 . Given that, 9 x 2 + 1 4 x 2 = 13 Add 2 3 x 1 2 x on both sides of the equation. ⇒ 9 x 2 + 1 4 x 2 + 2 3 x 1 2 x = 13 + 2 3 x 1 2 x It is in the form of a 2 + b 2 + 2 a b where a = 3 x and b = 1 2 x ⇒ 3 x 2 + 2 3 x 1 2 x + 1 2 x 2 = 13 + 3 Apply a 2 + 2 a b + b 2 = a + b 2 ⇒ 3 x + 1 2 x 2 = 16 ⇒ 3 x + 1 2 x = 4 Taking cube on both sides, 3 x + 1 2 x 3 = 4 3 Using identity, a + b 3 = a 3 + b 3 + 3 a b ( a + b ) 27 x 3 + 1 8 x 3 + 3 × 3 x × 1 2 x × 3 x + 1 2 x = 64 ⇒ 27 x 3 + 1 8 x 3 + 9 2 4 = 64 ⇒ 27 x 3 + 1 8 x 3 = 46 Hence, the value of 27 x 3 + 1 8 x 3 = 46

If 25 x 2 + 1 4 x 2 = 20 , find the value of 5 x + 1 2 x

We need to find the value of 5 x + 1 2 x . Given that, 25 x 2 + 1 4 x 2 = 20 Add 2 5 x 1 2 x on both sides of the equation. ⇒ 25 x 2 + 1 4 x 2 + 2 5 x 1 2 x = 20 + 2 5 x 1 2 x It is in the form of a 2 + b 2 + 2 a b where a = 5 x and b = 1 2 x ⇒ 5 x 2 + 2 5 x 1 2 x + 1 2 x 2 = 20 + 5 Apply a 2 + 2 a b + b 2 = a + b 2 ⇒ 5 x + 1 2 x 2 = 25 ⇒ 5 x + 1 2 x = ± 5 Hence, the value of 5 x + 1 2 x is ± 5 .

If 25 x 2 + 1 4 x 2 = 20 , find the value of 125 x 3 + 1 8 x 3

We need to find the value of 125 x 3 + 1 8 x 3 . Given that, 25 x 2 + 1 4 x 2 = 20 Add 2 5 x 1 2 x on both sides of the equation. ⇒ 25 x 2 + 1 4 x 2 + 2 5 x 1 2 x = 20 + 2 5 x 1 2 x It is in the form of a 2 + b 2 + 2 a b where a = 5 x and b = 1 2 x ⇒ 5 x 2 + 2 5 x 1 2 x + 1 2 x 2 = 20 + 5 Apply a 2 + 2 a b + b 2 = a + b 2 ⇒ 5 x + 1 2 x 2 = 25 ⇒ 5 x + 1 2 x = 5 Taking cube on both sides, 5 x + 1 2 x 3 = 5 3 Using identity, a + b 3 = a 3 + b 3 + 3 a b ( a + b ) 125 x 3 + 1 8 x 3 + 3 × 5 x × 1 2 x × 5 x + 1 2 x = 125 ⇒ 125 x 3 + 1 8 x 3 + 15 2 5 = 125 ⇒ 125 x 3 + 1 8 x 3 = 175 2 Hence the value of 125 x 3 + 1 8 x 3 = 175 2

E M B I B E

ICSE MATHEMATICS CLASS 9>Algebra>Chapter 3 - Expansions>EXERCISE 3B

B Nirmala Shastry Solutions for Exercise 2: EXERCISE 3B

Author:B Nirmala Shastry

B Nirmala Shastry Mathematics Solutions for Exercise - B Nirmala Shastry Solutions for Exercise 2: EXERCISE 3B

Attempt the practice questions from Exercise 2: EXERCISE 3B with hints and solutions to strengthen your understanding. ICSE MATHEMATICS CLASS 9 solutions are prepared by Experienced Embibe Experts.