Expanding Trinomials

Algebra Level 2

Find the sum of all (numerical) coefficients in the expansion of $(x+y+z)^{3}$ .

2 solutions

Akiva Weinberger
May 4, 2015

This is what you'd get if you set $x$ , $y$ , and $z$ to $1$ .

So the answer is $(1+1+1)^3=27$ .

Can you explain why we can set $x=y=z=1$ ? Why not $x=1,y=2,z=3$ ?

Bonus question : Can you find the number of terms in the trinomial expansion $(x+y+z)^3$ ? Can you generalize this for $(a_1+a_2+\ldots+a_n)^m$ ?

To Challenge Master: Multinomial Theorem?

Shashank Rammoorthy - 5 years, 11 months ago

Vinayak Srivastava
May 14, 2020

$(x+y+z)^{3} =x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^3$

Sum of coefficients $=1+1+1+3+3+3+3+3+3+6=3\times1+6\times3+1\times6 =3 + 18+ 6 =\boxed{27}$