Question:
Find the value
$x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}$
Solution:
$=(x)^{3}+(2 y)^{3}+3 \times x^{2} \times 2 y+3 \times x \times(2 y)^{2}$
$=(x+2 y)^{3}$
$\left[\therefore x^{3}+y^{3}+3 x^{2} y+3 x y^{2}=(x+y)^{3}\right]$
$=(x+2 y)(x+2 y)(x+2 y)$
$\therefore x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}=(x+2 y)(x+2 y)(x+2 y)$
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