Factorise of the following : $64 m^{3}-343 n^{3}$

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Using the identity $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right),$   we have 

$64 m ^{3}-343 n ^{3} =(4 m )^{3}-(7 n )^{3}=(4 m -7 n )\left[(4 m )^{2}+(4 m )(7 n )+(7 n )^{2}\right]$

$=(4 m -7 n )\left(16 m ^{2}+28 mn +49 n ^{2}\right)$

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