**(i) 9 – x ^{2 }+ 2xy – y^{2}**

**(ii) 9x ^{4} – (x^{2} + 2x + 1)**

**Answer :**

**(i) 9 – x**^{2 }+ 2xy – y^{2}

^{2 }+ 2xy – y

^{2}

9 – x^{2} + 2xy – y^{2}

Above terms can be written as,

9 – x^{2} + xy + xy – y^{2}

Now,

9 – x^{2} + xy + 3x – 3x + 3y – 3y + xy – y^{2}

Rearranging the above terms, we get,

9 – 3x + 3y + 3x – x^{2} + xy + xy – 3y – y^{2}

Take out common in all terms,

3(3 – x + y) + x(3 – x + y) + y (-3 – y + x)

3(3 – x + y) + x(3 – x + y) – y(3 – x + y)

(3 – x + y) (3 + x – y)

**(ii) 9x**^{4} – (x^{2} + 2x + 1)

^{4}– (x

^{2}+ 2x + 1)

9x^{4} – (x^{2} + 2x + 1)

Above terms can be written as,

(3x^{2})^{2} – (x + 1)^{2} … [because (a + b)^{2} = a^{2} + 2ab + b^{2}]

We know that, a^{2} – b^{2} = (a + b) (a – b)

So, (3x^{2} + x + 1) (3x^{2} – x – 1)

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