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    (2x^2 / 4y^3)^-2
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    (Original post by Tom.Dunwoody)
    (2x^2 / 4y^3)^-2
    Use (ab)^n = a^n b^n
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    Your expression can be written as (4y^3/2x^2)^2, which expanded then = 16y^6/4x^4. Simplifying this = 4y^6/x^4.
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    Sorry, made a small error before, fixed it.
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    (Original post by 2014FHD)
    Your expression can be written as (4y^3/2x^2)^2, which expanded then = 16y^6/4x^4. Simplifying this = 4y^6/x^4.
    how would you know to flip the expression round ? rather than (2x^2 / 4y^3) it is now (4y^3 / 2x^2)
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    (Original post by Tom.Dunwoody)
    how would you know to flip the expression round ? rather than (2x^2 / 4y^3) it is now (4y^3 / 2x^2)
    The initial equation says -2 outside the bracket. When the power is negative (e.g. x^-2), you take the reciprocal (1/x) for that number (therefore for a fraction, just flip the top and bottom number).

    After this, just keep the positive power (in this case 2) the same (so it is 1/x^2), and work from there
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    Okay so we have:

    (2X^2/4Y^3)^-2

    can be simplified to:

    (2/4 * X^2/Y^3)^-2

    further simplification:

    (X^2/2Y^3)^-2

    Therefore flipping the fraction due to the negative power:

    (2Y^3/X^2)^2

    Thus leaving you with:

    (4Y^6/X^4)

    Hope this helped.






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Updated: January 20, 2015
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