Answers

2015-07-11T10:58:30+08:00
Cube the first term .(9ab)(9ab)(9ab)=729a^3b^3
thrice the product of the square of first and last term
3(9ab)^2(16bc)=3 888ab^2c
thrice the product of the square of last and the first term
3(9ab)(16bc)^2=6 912ab^2c
cube the last term
(16bc)(16bc)(16bc)=4 096b^3c^3

=(729a^3b^3+3 888ab^2c+6 912ab^2c+4 096b^3c^3)

kanang ^3 ^2 kay exponent na..

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