Question

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Answer 1

Відповідь:

1

Пояснення:

$\frac{x^{4}+xy^{3} -x^{3}y-4^{4} }{(x^{2} -xy^{2} ) * (x^{2} -y^{2} )} = \frac{x * (x^{3}+y^{3} ) - y^{3} * (x^{3} - y^{3} ) }{(x^{2} -xy+y^{2} ) * (x-y) * (x+y)} = \frac{(x^{3} + y^{3} ) * (x-y)}{(x^{2} -xy+y^{2} ) * (x-y) * (x+y)} = \frac{(x^{2} - xy + y^{2} ) * 1 }{x^{2} -xy + y^{2} } = 1$

Answer 2

$\displaystyle\bf\\\frac{x^{4}+xy^{3} -x^{3}y-y^{4} }{(x^{2} -xy+y^{2})\cdot(x^{2} - y^{2}) } =\\\\\\=\frac{(x^{4} -x^{3}y)+(xy^{3} -y^{4}) }{(x^{2} -xy+y^{2})\cdot(x^{2} - y^{2}) } =\\\\\\=\frac{x^{3}\cdot(x -y)+y^{3}\cdot(x -y) }{(x^{2} -xy+y^{2})\cdot(x^{2} - y^{2}) } =\\\\\\=\frac{(x-y)\cdot(x^{3} +y^{3}) }{(x^{2} -xy+y^{2})\cdot(x - y)\cdot(x+y) } =\\\\\\=\frac{x^{3} +y^{3} }{(x^{2} -xy+y^{2})\cdot(x+y) } =\\\\\\=\frac{x^{3}+y^{3} }{x^{3}+y^{3} } =1\\\\\\Otvet \ : \ 1$