Question

you can also ask from Google please please this is the question ​

Answer 1

Step-by-step explanation:

${( \frac{ {5}^{a} }{ {5}^{b} })}^{a + b} \times {( \frac{ {5}^{b} }{ {5}^{c} })}^{b + c} \times {( \frac{ {5}^{c} }{ {5}^{a} })}^{a + c} = 1 \\ lhs = {( \frac{ {5}^{a} }{ {5}^{b} })}^{a + b} \times {( \frac{ {5}^{b} }{ {5}^{c} })}^{b + c} \times {( \frac{ {5}^{c} }{ {5}^{a} })}^{a + c} \\ lhs = ( {{5}^{a - b} }) ^{a + b} \times ( {{5}^{b - c} }) ^{b + c} \times( {{5}^{c - a} }) ^{c + a} \\lhs = {5}^{ {a}^{2} - {b}^{2} } \times {5}^{ {b}^{2} - {c}^{2} } \times {5}^{ {c}^{2} - {a}^{2} } \\ lhs = {5}^{{a}^{2} - {b}^{2} +{b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } = {5}^{0} = 1$

Hence Proved...