Given (abc+bca+cab)2
Using Identity,
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
∴(abc+bca+cab)2
=(abc)2+(bca)2+(cab)2+2×abc×bca+2×bca×cab+2×cab×abc
=(abc)2+(bca)2+(cab)2+2×1c2+2×1a2+2×1b2
=a2b2c2+b2c2a2+c2a2b2+2c2+2a2+2b2
Hence, (abc+bca+cab)2=a2b2c2+b2c2a2+c2a2b2+2c2+2a2+2b2.