Answer
$(10x+7y)(100x^{2}-70xy+49y^{2})$
Work Step by Step
$1000x^{3}=(10x)^{3},\qquad 343y^{3}=(7y)^{3}$
so we recognize this as a sum of cubes, $A^{3}+B^{3}=(A+B)(A^{2}-AB+B^{2})$
$...=(10x+7y)[(10x)^{2}-(10x)(7y)+(7y)^{2}]$
$=(10x+7y)(100x^{2}-70xy+49y^{2})$