Question:
Find the value
$64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$
Solution:
$=(4 a)^{3}+(5 b)^{3}+3(4 a)^{2}(5 b)+3(4 a)(5 b)^{2}$
$\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$
$=(4 a+5 b)^{3}$
$=(4 a+5 b)(4 a+5 b)(4 a+5 b)$
$\therefore 64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$
$=(4 a+5 b)(4 a+5 b)(4 a+5 b)$
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.