Answers

2014-09-12T23:11:14+05:30
[5y]³+[4z]³
by using identity a³+b³=a³+b³+3a²b+3ab²
[5y]³+[4z]³+3×[5y]²×[4z]+3×[5y]×[4z]²
125y³+64z³+225y²z+240yz²
0
2014-09-12T23:39:06+05:30