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Is there anybody who can help me with this proof? I know the basics, but can't figure this out.

Let $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ be vectors and $\boldsymbol{T}$ a tensor. Recall that the mixed product of the three vectors is denoted by $$\boldsymbol{a}\cdot(\boldsymbol{b} \times \boldsymbol{c}) = [\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}].$$ Prove that $$[\boldsymbol{Ta},\boldsymbol{Tb},\boldsymbol{Tc}] = \det(\boldsymbol{T})[\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}].$$ Provide a geometric interpretation.

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    Hi and welcome! For our questions here we use Mathjax, I've editted this question for you so that it is easier to read, but you can learn how to format your questions here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference Additionally, giving the attempts you've tried or approaches you've used help people give you better answers. Cheers! – 2017-02-12
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    hello ! I could not see the changes you suggested because i do not have the "reputation" needed. Could you see the picture that I attached or no ? – 2017-02-12
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    Have a look page 85 of the Google book "Fundamentals of Structural Mechanics" by Keith D. Hjelmstad – 2017-02-12

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How you prove this depends on where you are starting from. In particular, if you know $$\boldsymbol b \times \boldsymbol c = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=\mathbf i \begin{vmatrix}b_2&b_3\\c_2&c_3\end{vmatrix} + \mathbf j \begin{vmatrix}b_3&b_1\\c_3&c_1\end{vmatrix} + \mathbf k \begin{vmatrix}b_1&b_2\\c_1&c_2\end{vmatrix}$$then it becomes obvious that $$\boldsymbol a \cdot (\boldsymbol b \times \boldsymbol c) = a_1 \begin{vmatrix}b_2&b_3\\c_2&c_3\end{vmatrix} + a_2 \begin{vmatrix}b_3&b_1\\c_3&c_1\end{vmatrix} + a_3 \begin{vmatrix}b_1&b_2\\c_1&c_2\end{vmatrix}=\begin{vmatrix} a_1 & a_2 & a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}$$

Now determinants are unchanged by transposition, so we can also write $$\boldsymbol a \cdot (\boldsymbol b \times \boldsymbol c) =\begin{vmatrix} a_1 & b_1 & c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} = \det(\begin{bmatrix}\boldsymbol a&\boldsymbol b&\boldsymbol c\end{bmatrix})$$

(Note that the bracket notation on the right is not the triple product notation you wrote, but rather represents the matrix whose columns are the vectors $\boldsymbol a, \boldsymbol b,$ and $\boldsymbol c$.)

By the rules of matrix multiplication, $$\boldsymbol T\begin{bmatrix}\boldsymbol a&\boldsymbol b&\boldsymbol c\end{bmatrix} = \begin{bmatrix}\boldsymbol{Ta}&\boldsymbol{Tb}&\boldsymbol{Tc}\end{bmatrix}$$

And since the determinant is multiplicative, $$\begin{align}\boldsymbol{Ta} \cdot (\boldsymbol{Tb} \times \boldsymbol{Tc}) &= \det(\begin{bmatrix}\boldsymbol{Ta}&\boldsymbol{Tb}&\boldsymbol{Tc}\end{bmatrix})\\&=\det(\boldsymbol T\begin{bmatrix}\boldsymbol a&\boldsymbol b&\boldsymbol c\end{bmatrix})\\&=\det(\boldsymbol T)\det(\begin{bmatrix}\boldsymbol a&\boldsymbol b&\boldsymbol c\end{bmatrix})\\&=\det(\boldsymbol T)\boldsymbol (a \cdot (\boldsymbol b \times \boldsymbol c))\end{align}$$