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  1. #1
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    simplification

    Could someone explain in more details this simplification?
    Thanks
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  2. #2
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    Re: simplification

    Quote Originally Posted by cxz7410123 View Post
    Could someone explain in more details this simplification?
    Thanks
    Are you familiar with the difference of two squares? Namely a^2-b^2 = (a-b)(a+b)

    For a^4-b^4 use the same formula but with (a^2-b^2)(a^2+b^2). The first term can be simplified again using the difference of two squares.
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  3. #3
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    Re: simplification

    I hope you know the formula: a^2-b^2=(a-b)(a+b) (*).

    For the first fraction, the numerator is (a^2-b^2)^4, using (*) it becomes: ((a-b)(a+b))^4=(a-b)^4 \cdot (a+b)^4. (I used the rule (a \cdot b)^n=a^n \cdot b^n)

    For the second fraction, the numerator is (a^4-b^4), which can also be written: ((a^2)^2-(b^2)^2) (I used the rule (a^n)^m=a^{n \cdot m}). Using (*) again, we obtain ((a^2)^2-(b^2)^2)=(a^2-b^2)(a^2+b^2).


    So we have:
    \frac{(a-b)^4 \cdot (a+b)^4}{(a^2+b^2)} \cdot \frac{(a^2-b^2)\cdot(a^2+b^2)}{(a+b)^4}=\frac{(a-b)^4 \cdot (a+b)^4 \cdot (a^2-b^2) \cdot (a^2+b^2)}{(a^2+b^2)\cdot(a+b)^4}=\frac{(a^2+b^2) \cdot(a+b)^4 \cdot(a-b)^4 \cdot (a^2-b^2)}{(a^2+b^2)\cdot(a+b)^4}=(a-b)^4 \cdot (a^2-b^2)
    (I rearranged the factors, I hope this helped).


    Now, if we want we can apply (*) again for (a^2-b^2):
    (a-b)^4 \cdot (a^2-b^2)=(a-b)^4 \cdot (a-b)\cdot(a+b)= (a-b)^5\cdot(a+b) (I used the rule a^n \cdot a^m =a^{n+m}).
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