+0

# Binomial Theorem

0
276
2

In the expansion of (ab + bc + ca)^{12}, find the coefficient of a^8 b^8 c^8.

May 2, 2022

#1
+2

34,650 a^8 b^8 c^8

May 2, 2022
#2
+118652
+2

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}\\~\\

May 3, 2022