Show that if \(\alpha, \beta\), and \(\gamma\) are 1-forms, then \(\alpha \wedge \beta \wedge \gamma\) has components $$ \frac{1}{6}\left(\alpha_{a} \beta_{b} \gamma_{c}+\alpha_{b} \beta_{c} \gamma_{a}+\alpha_{c} \beta_{a} \gamma_{b}-\alpha_{a} \beta_{c} \gamma_{b}-\alpha_{b} \beta_{a} \gamma_{c}-\alpha_{c} \beta_{b} \gamma_{a}\right) $$

Question: If α, β, and γ are 1-forms, show that the components of their triple wedge product, α ∧ β ∧ γ , have the form: $ \frac{1}{6}\left(\alpha_{a} \beta_{b} \gamma_{c}+\alpha_{b} \beta_{c} \gamma_{a}+\alpha_{c} \beta_{a} \gamma_{b}-\alpha_{a} \beta_{c} \gamma_{b}-\alpha_{b} \beta_{a} \gamma_{c}-\alpha_{c} \beta_{b} \gamma_{a}\right). $ Solution: We've derived the components of α ∧ β ∧ γ for arbitrary 1-forms α, β, and γ and found it to be: $ \frac{1}{6}\left(\alpha_{a} \beta_{b} \gamma_{c}+\alpha_{b} \beta_{c} \gamma_{a}+\alpha_{c} \beta_{a} \gamma_{b}-\alpha_{a} \beta_{c} \gamma_{b}-\alpha_{b} \beta_{a} \gamma_{c}-\alpha_{c} \beta_{b} \gamma_{a}\right), $ which matches the desired expression.