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secondary 3 | A Maths
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① a³ - b³ = (a – b)(a² + ab + b²)
②a² - b² = (a - b)(a + b)
③(a + b)² = a² + 2ab + b²
So x^6 - 64
= (x²)³ - 4³
= (x² - 4)((x²)² + x²(4) + 4²) ( property ①)
= (x + 2)(x - 2)( (x²)² + 4x² + 4x² + 4² - 4x²) (property②)
= (x + 2)(x - 2)( (x²)² + 2(x²)(4) + 4² - 4x²)
(Completing the square)
= (x + 2)(x - 2)( (x² + 4)² - (2x)²)
(applying property ③)
= (x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x)
(Property ② again)
ii)
x^6 - 64 = (x² + 4)² - 4x²
So (x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) = (x² + 4)² - 4x²
(x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) = (x² + 4 - 2x)(x² + 4 + 2x)
(from i)
(x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) - (x² + 4 - 2x)(x² + 4 + 2x) = 0
( (x + 2)(x - 2) - 1)(x² + 4 - 2x)(x² + 4 + 2x) = 0
((x + 2)(x - 2) - 1) = 0 or x² + 4 - 2x = 0 or x² + 4 + 2x = 0
x² - 4 - 1 = 0 or x² + 4 - 2x = 0 or x² + 4 + 2x = 0
x² - 5 = 0 or x² - 2x + 1 = -3 or x² + 2x + 1 = -3
x² = 5 or (x - 1)² = -3 or (x + 1)² = -3
(Second and third case are rejected as (x - 1)² and (x + 1)² ≥ 0 for all real x. The square of a real value is either 0 or positive)
x = √5 or x = -√5
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