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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 $\displaystyle a^2-b^2 = (a-b)(a+b)$

    For $\displaystyle a^4-b^4$ use the same formula but with $\displaystyle (a^2-b^2)(a^2+b^2)$. 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: $\displaystyle a^2-b^2=(a-b)(a+b)$ (*).

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

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


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


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