question_answer

Prove that \[a\cdot (b+c)\times (a+2b+3c)=[\begin{matrix}    a & b & c  \\ \end{matrix}].\]

Answer:

We have,             \[a\cdot (b+c)\times (a+2b+3c)\] \[=a\cdot \{(b+c)\times (a+2b+3c)\}\]             \[=a\cdot \{b\times a+2(b\times b)+3(b\times c)\]                                      \[+c\times a+2(c\times b)+3(c\times c)\}\] \[=a\cdot \{b\times a+3(b\times c)+c\,\times a-2(b\times c)\}\]             \[=a\cdot \{-(a\times b)+b\times c+c\times a\}\]             \[=-a\cdot (a\times b)+a\cdot (b\times c)+a\cdot (c\times a)\]             \[=0+[a\,\,b\,\,c]+0\]             \[=[a\,\,b\,\,c]\]