math
posted by Roy Jr .
(2x+3y+4z) to the 3rd power

There is no easy rule, because there are many terms.
The following identity could help:
(a+b+c)^3
=c^3+3bc^2+3ac^2+3b^2c+6abc+3a^2c+b^3+3ab^2+3a^2b+a^3
So substitute 2x for a, 3y for b and 4z for c. Do not forget that a^2 would mean (2x)^2=4x^2, a^3 would mean (2x)^3=8a^3 and so on.
A good way to check your work is to add up the coefficients, they should equal (2+3+4)^3 = 729. 
very patiently expand
(2x + 3y + 4z)(2x + 3y + 4z)(2x + 3y + 4z)
= ....
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