Question:
Find the value
$8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}$
Solution:
$=(2 x)^{3}+(y)^{3}+3 \times(2 x)^{2} \times y+3(2 x) \times y^{2}$
$=(2 x+y)^{3}$
$\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$
$=(2 x+y)(2 x+y)(2 x+y)$
$\therefore 8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}=(2 x+y)(2 x+y)(2 x+y)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.