simplification

**cxz7410123** · November 11th 2011, 08:07 AM

Could someone explain in more details this simplification?
Thanks

**e^(i*pi)** · November 11th 2011, 08:12 AM

Originally Posted by cxz7410123

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.

**veileen** · November 11th 2011, 08:28 AM

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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