Number of terms in trinomial expansion

Question

We know that the number of terms in the expansion of $(x_1+x_2+\cdots+x_k)^n$ is $\ ^{n+k-1}\mathrm{C}_{k-1}$. Using that formula,the number of terms in $(a^2+2ab+b^2)^3$ should be $\ ^{3+2}C_2$ or $\ ^5C_2=10$. But if we notice,then $(a^2+2ab+b^2)^3=(a+b)^6$. So,the number of terms should be $6+1=7$. Where is the fault with using the general formula then?

Answer 1

You said , the number of terms in $(a^2+2ab+b^2)^3$ and $(a+b)^6$ are 10 and 7 respectively.

It is true that : $(a^2+2ab+b^2)^3$ = $(a+b)^6$ .

However, it is not necessary their each terms are equal after expansion. They will add upto and can be simplified to give same expressions after expansion, however their each terms in the first expansion are not supposed to be equal.

Consider a different type of example, the less-simplified expansion of $(x)^2$ .

$[(x-1)+1]^2 = (x-1)^2 + 2(x-1) + 1$

Also, we know that $x^2 = (x)^2$

However, you cannot compare the number of terms in them because individual terms will be different while their value will always be same.