Question
Show that:$\Big(\frac{3^\text{a}}{2^{\text{b}}}\Big)^{\text{a}+\text{b}}\Big(\frac{3^\text{b}}{3^\text{c}}\Big)^{\text{b}+\text{c}}\Big(\frac{3^\text{c}}{3^\text{a}}\Big)^{\text{c}+\text{a}}=1$
✓
Answer
$\text{LHS}=\Big(\frac{3^\text{a}}{2^{\text{b}}}\Big)^{\text{a}+\text{b}}\Big(\frac{3^\text{b}}{3^\text{c}}\Big)^{\text{b}+\text{c}}\Big(\frac{3^\text{c}}{3^\text{a}}\Big)^{\text{c}+\text{a}}$$=\big(3^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(3^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(3^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}$
$=\big(3^{(\text{a}-\text{b})(\text{a}+\text{b})}\big)\big(3^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(3^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(3^{\text{a}^2-\text{b}^2}\big)\big(3^{\text{b}^2-\text{c}^2}\big)\big(3^{\text{c}^2-\text{a}^2}\big)$
$=3^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=3^0$
$=1$
$=\text{RHS}$
$=\big(3^{(\text{a}-\text{b})(\text{a}+\text{b})}\big)\big(3^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(3^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(3^{\text{a}^2-\text{b}^2}\big)\big(3^{\text{b}^2-\text{c}^2}\big)\big(3^{\text{c}^2-\text{a}^2}\big)$
$=3^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=3^0$
$=1$
$=\text{RHS}$
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