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Substitution Method to Remove Indeterminate Form

Prove that: a...

Question

Prove that: $(\vec{a} - \vec{b}) \cdot \{(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})\} = 0$

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Solution

$We have (\vec{a} - \vec{b}) . \{(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})\} = (\vec{a} - \vec{b}) . (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}) (By distributive law) = (\vec{a} - \vec{b}) . (\bar{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) (∵ \vec{c} \times \vec{c} = 0) = (\vec{a} - \vec{b}) . (\bar{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}) = \vec{a} . (\vec{b} \times \vec{c}) + \vec{a} . (\overset{⇀}{a} \times \vec{b}) + \vec{a} . (\vec{c} \times \vec{a}) - \vec{b} . (\vec{b} \times \vec{c}) - \vec{b} . (\vec{a} \times \vec{b}) - \vec{b} . (\vec{c} \times \vec{a}) (By distributive law) = [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] + [\begin{matrix} \vec{a} & \vec{a} & \vec{b} \end{matrix}] + [\begin{matrix} \vec{a} & \vec{c} & \vec{a} \end{matrix}] - [\begin{matrix} \vec{b} & \vec{b} & \vec{c} \end{matrix}] - [\vec{b} \vec{a} \vec{b}] - [\begin{matrix} \vec{b} & \vec{c} & \vec{a} \end{matrix}] = [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] + 0 + 0 - 0 - 0 - [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] (∵ [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] = [\begin{matrix} \vec{b} & \vec{c} & \vec{a} \end{matrix}]) = 0$

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Q. Prove that :

x^{a - b} \times x^{b - c} \times x^{c - a} = 1

Q. Prove that :

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + a^{a - c}} = 1

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

Q. Prove that:

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\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

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