Resolve into factors: $m^{4} + n^{4} - 18 m^{2} n^{2}$

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Solution

In the given expression $m^{4} + n^{4} - 18 m^{2} n^{2}$ , add and subtract $2 m^{2} n^{2}$ to make it a perfect square as shown below:

$m^{4} + n^{4} - 18 m^{2} n^{2} = (m^{4} + n^{4} + 2 m^{2} n^{2}) - 18 m^{2} n^{2} - 2 m^{2} n^{2} = (m^{2} + n^{2})^{2} - 20 m^{2} n^{2}$
(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 $(m^{2} + n^{2})^{2} - 20 m^{2} n^{2}$ can be factorised as follows:

$(m^{2} + n^{2})^{2} - 20 m^{2} n^{2} = (m^{2} + n^{2})^{2} - (\sqrt{20} m n)^{2} = (m^{2} + n^{2} + \sqrt{20} m n) (m^{2} + n^{2} - \sqrt{20} m n)$

Hence, $m^{4} + n^{4} - 18 m^{2} n^{2} = (m^{2} + n^{2} + \sqrt{20} m n) (m^{2} + n^{2} - \sqrt{20} m n)$

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