using multinomial theorem to expand ($2x -y + 3z)^3$

Question

I know how to set it up and there is for this example $10$ terms. But what is the best way to find the expanded work plus finding the final answer?

Sorry if there is a duplicate to this problem. I searched but couldn't find any with the information I desired.

Answer 1

the options for the exponents are: $(3,0,0),(2,1,0),(2,0,1),(1,2,0),(1,1,1),(1,0,2),(0,3,0),(0,2,1),(0,1,2),(0,0,3)$

Now use the multinomial theorem to figure out the respective multinomial coefficients which should be

$1,3,3,3,6,3,1,3,3,1$ in that order. Now expand using everything to get:

$(2x-y+3z)^3=(2x)^3+\mathbf3(2x)^2(-y)+\mathbf3(2x)^2(3z)+\mathbf3(2x)(-y)^2+\mathbf6(2x)(-y)(3z)+\mathbf3(2x)(3z)^2+(-y)^3+\mathbf3(-y)^2(3z)+\mathbf3(-y)(3z)^2+(3z)^3$

you can now leave this in a nicer way by doing the exponents and multiplying the coefficients, you should get:

$8x^3-12x^2y+36x^2z+6xy^2-36xyz+54xz^2-y^3+9y^ 2z -27yz^2+27z^3$

Answer 2

Have you tried Pascal's Triangle?

Also, awesome username.

$$ 1 $$ $$ 1 \space\space\space 1 $$ $$ 1 \space\space\space 2 \space\space\space 1 $$ $$ 1 \space\space\space 3 \space\space\space 3 \space\space\space 1 $$

$$ (2x - y)^3 + 3(2x - y)^23z + 3(2x - y)3^2z^2 + 3^3z^3$$ $$ (2x - y)^3 + 9(4x^2 - 4xy + y^2)z + 27(2x - y)z^2 + 27z^3$$

Now working on the $(2x - y)^3$ term, you can expand it out using the normal binomial.

Of course, you can always use the trinomial expansion,

$$ (a + b + c)^n = \sum_{i,j,k} \binom{n}{i,j,k}a^ib^jc^k .$$

For this you use, $\binom{n}{i,j,k} = \frac{n!}{i!j!k!} $ or Pascal's Pyramid. Although, this is kind of unnecessary.