Question
Show that: $\big(\text{x}^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(\text{x}^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(\text{x}^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}=1$
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Answer
$\text{LHS}=\big(\text{x}^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(\text{x}^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(\text{x}^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}$
$=\big(\text{x}^{(\text{a}+\text{b})(\text{a}+\text{b})}\big)\big(\text{x}^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(\text{x}^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(\text{x}^{\text{a}^2-\text{b}^2}\big)\big(\text{x}^{\text{b}^2-\text{c}^2}\big)\big(\text{x}^{\text{c}^2-\text{a}^2}\big)$
$=\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=\text{x}^0$
$=1$
$=\text{RHS}$
$=\big(\text{x}^{(\text{a}+\text{b})(\text{a}+\text{b})}\big)\big(\text{x}^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(\text{x}^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(\text{x}^{\text{a}^2-\text{b}^2}\big)\big(\text{x}^{\text{b}^2-\text{c}^2}\big)\big(\text{x}^{\text{c}^2-\text{a}^2}\big)$
$=\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=\text{x}^0$
$=1$
$=\text{RHS}$
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