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Math Help - Difference of Squares

  1. #1
    Member cmf0106's Avatar
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    Difference of Squares

    In the difference of squares it can come in the form of x^n-c^n where n is an even number. The factorization is x^n-c^n=(x^{\frac{n}{2}}-c^{\frac{n}{2}})(x^{\frac{n}{2}}+c^{\frac{n}{2}})

    However in the following problem I am lost at a particular step. Namely, how \frac{1}{64} turns into (\frac{1}{2})^6


    x^6-\frac{1}{64}=x^6-(\frac{1}{2})^6=(x^3-\frac{1}{2})(x^3+\frac{1}{2})
    If someone could walk me through this it would be greatly appreciated,
    Many Thanks
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    Note that 2^6 = 64 and the property of powers: \frac{a^{n}}{b^{n}} = \left(\frac{a}{b}\right)^n

    So: \frac{1}{64} = \frac{1^6}{2^6} = \left(\frac{1}{2}\right)^6
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    Quote Originally Posted by cmf0106 View Post
    In the difference of squares it can come in the form of x^n-c^n where n is an even number. The factorization is x^n-c^n=(x^{\frac{n}{2}}-c^{\frac{n}{2}})(x^{\frac{n}{2}}+c^{\frac{n}{2}})

    However in the following problem I am lost at a particular step. Namely, how \frac{1}{64} turns into (\frac{1}{2})^6


    x^6-\frac{1}{64}=x^6-(\frac{1}{2})^6=(x^3-\frac{1}{2})(x^3+\frac{1}{2})
    If someone could walk me through this it would be greatly appreciated,
    Many Thanks
    no big deal here. using the rules to distribute the exponent through a rational function, \left( \frac 12 \right)^6 = \frac {1^6}{2^6} = \frac 1{64} .....since \left( \frac xy \right)^a = \frac {x^a}{y^a}
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    Thank you very much guys. Honestly, in the whole exponent chapter this rule was never stated believe it or not

    Also once we established x^6-(\frac{1}{2})^6 why is it equal to (x^3-\frac{1}{2})(x^3+\frac{1}{2}) shouldnt the 6 exponent be divided by 2 resulting in (x^3-\frac{1}{2}^3)(x^3+\frac{1}{2}^3)
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    Quote Originally Posted by cmf0106 View Post
    Thank you very much guys. Honestly, in the whole exponent chapter this rule was never stated believe it or not

    Also once we established x^6-(\frac{1}{2})^6 why is it equal to (x^3-\frac{1}{2})(x^3+\frac{1}{2}) shouldnt the 6 exponent be divided by 2 resulting in (x^3-\frac{1}{2}^3)(x^3+\frac{1}{2}^3)
    (x^3-\frac{1}{8})(x^3+\frac{1}{8})

    Now you have two factors. The first factor is the difference of two cubes and the second factor is the sum of two cubes. Each of these factors can be factored further.
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    so irrespective of the fact the book did not factor completely would (x^3-\frac{1}{2})(x^3-\frac{1}{2}) be correct? If so how do you arrive at that particular answer

    Also could someone walk me through this problem as well 16x^4-1=(2x)^4-1^4=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)
    Lost in this one as well unfortunately . I understand breaking down 16x^4-1=(2x)^4-1^4 but how that yields (4x^2-1)(4x^2+1) I am not certain.
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  7. #7
    A riddle wrapped in an enigma
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    Quote Originally Posted by cmf0106 View Post
    so irrespective of the fact the book did not factor completely would (x^3-\frac{1}{{\color{red}8}})(x^3{\color{red}+}\frac{1  }{{\color{red}8}}) be correct? If so how do you arrive at that particular answer? To continue, you would have to know the model for the difference and sum of cubes. Is that part of your lesson?

    Also could someone walk me through this problem as well 16x^4-1=(2x)^4-1^4=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)
    Lost in this one as well unfortunately . I understand breaking down 16x^4-1=(2x)^4-1^4 but how that yields (4x^2-1)(4x^2+1) I am not certain.
    Maybe you could look at it this way:

    16x^4-1=(4x^2)^2-(1^2)^2 which now looks like the difference between two squares.
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    Quote Originally Posted by masters View Post
    Maybe you could look at it this way:

    16x^4-1=(4x^2)^2-(1^2)^2 which now looks like the difference between two squares.
    With all due respect Masters, im afraid that does not help. Its most likely because of my own ignorance, but this book does tend to skip around alot.
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    A riddle wrapped in an enigma
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    Quote Originally Posted by cmf0106 View Post
    With all due respect Masters, im afraid that does not help. Its most likely because of my own ignorance, but this book does tend to skip around alot.
    16x^4-1=(4x^2)^2-(1^2)^2

    Let u=4x^2 and v=1^2

    u^2-v^2=(u-v)(u+v)

    Now substutute back in for u and v:

    (4x^2-1)(4x^2+1)

    Remember 1 raised to any power is 1.
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    This is a handy site for understanding the factoring of the difference of two squares and also has links for the sum and difference of cubes as well:

    Special Factoring: Differences of Squares
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