level: LEVEL 2 ALGEBRAIC IDENTITY 6 & 7
Questions and Answers List
Find the cube of each of the following binomial expressions)
level questions: LEVEL 2 ALGEBRAIC IDENTITY 6 & 7
| Question | Answer |
|---|---|
| Factorize the given expression: 27x³ - 63x²+ 49x - 343/27 | Compare the given expression to the standard cubic algebraic identity to factorise (a - b)3 = a3 - 3a2b + 3ab2 - b3 Compare 27x3 - 63x2 + 49x - 34327 to a3 - 3a2b + 3ab2 - b3 We can deduce that 27x3 = a3 or (3x)3 = a3 ∴ a = 3x And 34327 = b3 or [73]3 = b3 ∴ b = 73 So, 27x3 - 63x2 + 49x - 34327 |
| (i) (1/x + y/3) | [Using identities: (a + b)³ = a³ + b³ + 3ab(a + b) and (a – b)³ = a³ – b³ – 3ab(a – b) ] |
| (ii) (x/2 + y/3) ³ | (x/2 + y/3) ³ [Using identities: (a + b)³ = a³ + b³ + 3ab(a + b) and (a – b)³ = a³ – b³ – 3ab(a – b) Here a = (x/2 ) and b = ( y/3) x³/8 + y³ / 27 + 1/4 (3x²y + 2xy²) |
| Factorize the given expression: 27x³ - 63x²+ 49x - 343/27 | Compare the given expression to the standard cubic algebraic identity to factorise (a - b)³ = a³ - 3a²b + 3ab² - b³ Compare 27x³ - 63x² + 49x - 343/27 to a³ - 3a²b + 3ab² - b³ We can deduce that 27x³ = a³ or (3x)³ = a³ ∴ a = 3x And 343/27 = b³ or [73]³ = b³ ∴ b = 7/3 So, 27x³ - 63x²+ 49x - 343/27 (3x)³ - 3(3x)²(7/3) + 3(3x)[7/3]² - [7/3]³ = [3x−7/3]³ = [3x−7/3] [3x−7/3] [3x−7/3] |
| Using suitable identity evaluate the following: 98³ | (98)³ = (100 - 2)³ = (100 - 2)³is of the form (a - b)³ (a - b)³ = a³ - 3a²b + 3ab² - b³ So, (100 - 2)³ = 100³ - (3 × 100² × 2) + (3 × 100 × 2²) - 2³ = 1,000,000 - 60,000 + 1200 - 8 = 9,41,192 |