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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?

insipidintegrator
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madness
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    The formula does not allow for coincidences like $4 x_1 x_3=x_2^2$. – ancient mathematician Sep 03 '22 at 14:12
  • Then in which cases will the general formula be not applicable?Could you please tell me method to identify when we can't use the formula? – madness Sep 03 '22 at 14:21
  • method to identify when we can't use the formula --- There is no "general method". If there are dependence relations among the terms, then this has to be accounted for somehow on a case-by-case basis. The formula simply tells you how many terms occur for a certain expansion (e.g. $3x_1x_2^3 + 2x_2^3x_1$ is collapsed to $5x_1x_2^3,$ but otherwise the $x_k$'s are treated as independent variables), which might not be the expansion you want (e.g. one in which $4x_1x_2 + x_2^2$ is collapsed to $5x_2^2).$ – Dave L. Renfro Sep 03 '22 at 14:30

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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.

An_Elephant
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