Factorise: 4 x 2 + 1 4 x 2 + 2 - 9 y 2

Given expression: 4 x 2 + 1 4 x 2 + 2 - 9 y 2 By using the identity: x 2 - y 2 = x + y x - y 2 x + 1 2 x 2 and 9 y 2 = 3 y 2 = 2 x + 1 2 x 2 - 3 y 2 = 2 x 1 + 1 2 x + 3 y 1 2 x 1 + 1 2 x - 3 y 1 = 2 x + 1 2 x + 3 y 2 x + 1 2 x - 3 y

If x + 1 x = 7 , then find the value of x 4 + 1 x 4 .

Given expression: x + 1 x = 7 Find the value of x 4 + 1 x 4 As we know that x + 1 x 2 = 7 2 By using identity a + b 2 = a 2 + b 2 + 2 a b x 2 + 1 x 2 + 2 × x × 1 x = 49 x 2 + 1 x 2 + 2 = 49 x 2 + 1 x 2 = 47 x 2 + 1 x 2 2 = 47 2 x 4 + 1 x 4 + 2 × x 2 + 1 x 2 = 2209 x 4 + 1 x 4 = 2209 - 2 x 4 + 1 x 4 = 2207

If x + 1 x = 5 , then find the value of x 3 + 1 x 3 .

Given expression.: x + 1 x = 5 To find: x 3 + 1 x 3 According to question: x + 1 x 3 = 5 3 By using identity a + b 3 = a 3 + b 3 + 3 a b a + b x 3 + 1 x 3 + 3 × x × 1 x x + 1 x = 125 x 3 + 1 x 3 + 3 × 5 = 125 ∵ x + 1 x = 5 x 3 + 1 x 3 = 125 - 15 x 3 + 1 x 3 = 110

If a + b + c = 0 , then show that a 2 b c + b 2 c a + c 2 a b = 3 .

Given expression: a + b + c = 0 By using the identity if x + y + z = 0 t h e n x 3 + y 3 + z 3 = 3 x y z We know if a + b + c = 0 , a 3 + b 3 + c 3 = 3 a b c Dividing whole equation by a b c a 3 a b c + b 3 a b c + c 3 a b c = 3 a b c a b c ⇒ a 2 b c + b 2 ac + c 2 a b = 3

E M B I B E

Full Marks Mathematics 9>Chapter 2 - Polynomials>TEST YOUR SKILLS

R Chandra Solutions for Chapter: Polynomials, Exercise 6: TEST YOUR SKILLS

Author:R Chandra

R Chandra Mathematics Solutions for Exercise - R Chandra Solutions for Chapter: Polynomials, Exercise 6: TEST YOUR SKILLS

Attempt the practice questions on Chapter 2: Polynomials, Exercise 6: TEST YOUR SKILLS with hints and solutions to strengthen your understanding. Full Marks Mathematics 9 solutions are prepared by Experienced Embibe Experts.