Factorise: $4 x^{4} + 81 y^{4}$

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Solution

In the given equation $4 x^{4} + 81 y^{4}$ , add and subtract $36 x^{2} y^{2}$ to make it a perfect square as shown below:

$4 x^{4} + 81 y^{4} = (4 x^{4} + 81 y^{4} + 36 x^{2} y^{2}) - 36 x^{2} y^{2} = [(2 x^{2})^{2} + (9 y^{2})^{2} + 2 (2 x^{2}) (9 y^{2})] - 36 x^{2} y^{2} = (2 x^{2} + 9 y^{2})^{2} - 36 x^{2} y^{2}$
(using the identity $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ )

We also know the identity $a^{2} - b^{2} = (a + b) (a - b)$ , therefore,

Using the above identity, the equation $(2 x^{2} + 9 y^{2})^{2} - 36 x^{2} y^{2}$ can be factorised as follows:

$(2 x^{2} + 9 y^{2})^{2} - 36 x^{2} y^{2} = (2 x^{2} + 9 y^{2})^{2} - (6 x y)^{2} = (2 x^{2} + 9 y^{2} + 6 x y) (2 x^{2} + 9 y^{2} - 6 x y)$

Hence, $4 x^{4} + 81 y^{4} = (2 x^{2} + 9 y^{2} + 6 x y) (2 x^{2} + 9 y^{2} - 6 x y)$

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