Q: How do you factor 8 minus 27x cubed?

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27X^3-8 = 0 27X^3 = 8 X^3 = 8/27 X = cubic root of 8/cubic root of 27 = 2/3

(x3 - 8) is factored thus: (x - 2)(x2 + 2x + 4) The easiest way to do this is to remember the formula: (a3 - b3) = (a - b)(a2 + ab + b2)

(3x + 8)(3x - 8)

(2x + 5)(4x^2 - 10x + 25)

Answer: (x - 2)(x² + 2x + 4)Factor the difference of cubes:a³ - b³ = (a - b)(a² + ab + b² )note 8 = 2³ = 8 ⇒ b = 2 , with a = xSox³ - 8= x³ - 2³= (x - 2)(x² + 2x + 2²) = (x - 2)(x² + 2x + 4)

Related questions

27X^3-8 = 0 27X^3 = 8 X^3 = 8/27 X = cubic root of 8/cubic root of 27 = 2/3

the answer is (3x-2)(9x squared+6x+4)

20x2 - 27x -8 20 * -8 is -160, and -32 and 5 add to -27 20x2 - 32x + 5x - 8 4x(5x-8) + 1(5x-8) (4x + 1)(5x-8)

I also need to know the answer to this problem. Can anyone answer it?

8(y - 1)(y^2 + y + 1)

8(y - 1)(y^2 + y + 1)

All that is factorable here is the common factor t.t3 - 8tt(t2 - 8)======

9 minus 8

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(x + 1)(x - 8)

0.2963

x 3-8