In the expansion of (ab + bc + ca)^{12}, find the coefficient of a^8 b^8 c^8.
+118690 n the expansion of (ab + bc + ca)^{12}, find the coefficient of a^8 b^8 c^8.
\(L[a(b+c)+bc]^{12}\\~\\ \displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* [a(b+c)]^{r}\\~\\ \displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r(b+c)^{r}\\~\\ \displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r\left [\displaystyle \sum_{t=0}^r\binom{r}{t}b^rc^{t-r}\right]\\~\\ \displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r\left [\displaystyle \sum_{t=0}^r\binom{r}{t}b^tc^{r-t}\right]\\~\\ \)
Now I am only interested in the terms where abc are all to the power of 8 so r=8
\(\displaystyle \binom{12}{8}(bc)^{12-8}* a^8\left [\displaystyle \sum_{t=0}^8\binom{8}{t}b^tc^{8-t}\right]\\~\\ \displaystyle \binom{12}{8}b^4c^4* a^8\left [\displaystyle \sum_{t=0}^8\binom{8}{t}b^tc^{8-t}\right]\\~\\ So\;\; we\;\; need \;t=4\\~\\ \displaystyle \binom{12}{8}b^4c^4* a^8 \binom{8}{4}b^4c^{4}\\~\\ \displaystyle \binom{12}{8}\binom{8}{4} a^8 b^8c^{8}\\~\\\)
The coefficiant will be 495*70 = 34650
just as guest already told you. 
L[a(b+c)+bc]^{12}\\~\\
\displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* [a(b+c)]^{r}\\~\\
\displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r(b+c)^{r}\\~\\
\displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r\left [\displaystyle \sum_{t=0}^r\binom{r}{t}b^rc^{t-r}\right]\\~\\
\displaystyle \sum_{r=0}^{12}\binom{12}{r}(bc)^{12-r}* a^r\left [\displaystyle \sum_{t=0}^r\binom{r}{t}b^tc^{r-t}\right]\\~\\
\displaystyle \binom{12}{8}(bc)^{12-8}* a^8\left [\displaystyle \sum_{t=0}^8\binom{8}{t}b^tc^{8-t}\right]\\~\\
\displaystyle \binom{12}{8}b^4c^4* a^8\left [\displaystyle \sum_{t=0}^8\binom{8}{t}b^tc^{8-t}\right]\\~\\
So\;\; we\;\; need \;t=4\\~\\
\displaystyle \binom{12}{8}b^4c^4* a^8 \binom{8}{4}b^4c^{4}\\~\\
\displaystyle \binom{12}{8}\binom{8}{4} a^8 b^8c^{8}\\~\\
