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    Can someone show me a method of how to factorise x^3-8
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    Well, a good place to begin is to ask yourself if you can see any values of x for which x^3-8 = 0.
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    Look for values of x^3 = 8
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    Solve or factorize?
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    Solving x^3 - 8 = 0 and factorising x^3 - 8 are nearly the same thing
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    A to an odd power minus B to an odd power has a standard factorisation.
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    (Original post by Srai)
    x^3 - 8

    = (x - 2)^3
    No it isn't, unless x equals 2.
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    (Original post by stevencarrwork)
    No it isn't, unless x equals 2.
    Sorry, my excuse is that I haven't slept at all today.

    to make up for myself, the method you need to factorise x^3 - 8

    is the difference of two cubes.

    i.e.

    a^3 ± b^3 = (a ± b)(a^2 ∓ ab + b^2)

    so..

    x^3 - 8 = (x)^3 - (2)^3

    (x-2)(x^2 + 2x + 4)

    if you want test it.

    x^3 + 2x^2 + 4x - 2x^2 - 4x - 8

    = x^3 - 8

    try x^3 + 27
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    (Original post by Srai)
    Sorry, my excuse is that I haven't slept at all today.

    to make up for myself, the method you need to factorise x^3 - 8

    is the difference of two cubes.

    i.e.

    a^3 ± b^3 = (a ± b)(a^2 ∓ ab + b^2)

    so..

    x^3 - 8 = (x)^3 - (2)^3

    (x-2)(x^2 + 2x + 4)

    if you want test it.

    x^3 + 2x^2 + 4x - 2x^2 - 4x - 8

    = x^3 - 8

    try x^3 + 27
    thanks srai , i actually never knew you could have the difference of two cubes
 
 
 

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