Posted by **Roy Jr** on Wednesday, December 14, 2011 at 10:27pm.

(2x+3y+4z) to the 3rd power

- math -
**MathMate**, Wednesday, December 14, 2011 at 11:39pm
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.

- math -
**Reiny**, Wednesday, December 14, 2011 at 11:39pm
very patiently expand

(2x + 3y + 4z)(2x + 3y + 4z)(2x + 3y + 4z)

= ....

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