Solving Symmetric Tensor: c\cdot (A \times b) \neq (A \times b) \cdot c

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Homework Statement



Demostrate:
c\cdot (A \times b) \neq (A \times b) \cdot c

with A \in\Re^{3 \times 3} is a symmetric Tensor of second order and b,c \in \Re^3 are vectors

Homework Equations





The Attempt at a Solution



(A \times b)_ {ij} = A_{ij} \epsilon _{jkl} b_l
 
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Let's define another tensor T by T_{ij}=(A \times b)_ {ij} = A_{ij} \epsilon _{jkl} b_l...
what is c \cdot T ?...how about T \cdot c ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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