How to factorise x^6-64?
回答 (6)
2016-06-24 5:17 pm
✔ 最佳答案
x⁶ - 64 = (x³)² - (2³)²= (x³+2³) (x³-2³)
= ((x+2) (x²-2x+2²)) ((x-2) (x²+2x+2²))
= (x+2) (x-2) (x²-2x+4) (x²+2x+4)
= (x+2) (x-2) ((x²+4)-2x) ((x²+4)+2x)
2016-06-24 12:22 pm
Identities used :
a² - b² = (a + b)(a - b)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ = (a + b)(a² - ab + b²)
x⁶ - 64
= (x³)² - 8²
= (x³ + 8)(x³ - 8)
= (x³ + 2²)(x³ - 2²)
= (x + 2)(x² - 2x + 2²)(x + 2)(x² + 2x + 2²)
= (x + 2)(x - 2)(x² + 2x + 4)(x² - 2x + 4)
a² - b² = (a + b)(a - b)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ = (a + b)(a² - ab + b²)
x⁶ - 64
= (x³)² - 8²
= (x³ + 8)(x³ - 8)
= (x³ + 2²)(x³ - 2²)
= (x + 2)(x² - 2x + 2²)(x + 2)(x² + 2x + 2²)
= (x + 2)(x - 2)(x² + 2x + 4)(x² - 2x + 4)
2016-06-24 12:31 pm
Taking the low hanging fruit first, this is a difference of two squares
x^6 – 64 = (x^3 – 8)(x^3 + 8)
The difference of two cubes is used less often, but should be known. It is :-
a^3 – b^3 = (a – b)(a^2 + ab + b^2) …………….(1)
x^3 – 2^3 = (x – 2)(x^2 + 2x + 4) …………………..(2)
It is no problem to replace b by –b in (1) and apply it similarly to (2)
x^3 + 2^3 = (x + 2)(x^2 - 2x + 4) …………………..(2)
x^6 – 64 = (x – 2)(x + 2)(x^2 + 2x + 4)(x^2 - 2x + 4)
x^6 – 64 = (x^3 – 8)(x^3 + 8)
The difference of two cubes is used less often, but should be known. It is :-
a^3 – b^3 = (a – b)(a^2 + ab + b^2) …………….(1)
x^3 – 2^3 = (x – 2)(x^2 + 2x + 4) …………………..(2)
It is no problem to replace b by –b in (1) and apply it similarly to (2)
x^3 + 2^3 = (x + 2)(x^2 - 2x + 4) …………………..(2)
x^6 – 64 = (x – 2)(x + 2)(x^2 + 2x + 4)(x^2 - 2x + 4)
2016-06-24 12:09 pm
Well,
x^6 - 64 = ( (x^2)^3 - 4^3)
= (x^2 - 2^2)(x^4 + 4x^2 + 16)
= (x - 2)(x + 2)(x^4 + 4x^2 + 16)
hope it' ll help !!
x^6 - 64 = ( (x^2)^3 - 4^3)
= (x^2 - 2^2)(x^4 + 4x^2 + 16)
= (x - 2)(x + 2)(x^4 + 4x^2 + 16)
hope it' ll help !!
2016-06-24 1:38 pm
x⁶ - 64
= (x³)² - (2³)² : using [ a² - b² = (a + b)(a - b) ]
= (x³ + 2³)(x³ - 2³) : using [ a³ - b³ = (a - b)(a² + ab + b²) &
........................................... a³ + b³ = (a + b)(a² -ab + b²) ]
= (x + 2)(x² - 2x + 2²)(x - 2)(x² + 2x + 2²)
= (x + 2)(x - 2)(x² + 2x + 4)(x² - 2x + 4)
a² - b² = (a + b)(a - b)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ = (a + b)(a² -ab + b²)
= (x³)² - (2³)² : using [ a² - b² = (a + b)(a - b) ]
= (x³ + 2³)(x³ - 2³) : using [ a³ - b³ = (a - b)(a² + ab + b²) &
........................................... a³ + b³ = (a + b)(a² -ab + b²) ]
= (x + 2)(x² - 2x + 2²)(x - 2)(x² + 2x + 2²)
= (x + 2)(x - 2)(x² + 2x + 4)(x² - 2x + 4)
a² - b² = (a + b)(a - b)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ = (a + b)(a² -ab + b²)
2016-06-24 2:17 pm
x^6 - 64
= x^6 - 2^6
= (x - 2)(x + 2)(x^2 - 2x + 4)(x^2 + 2x + 4)
= x^6 - 2^6
= (x - 2)(x + 2)(x^2 - 2x + 4)(x^2 + 2x + 4)
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