Question

x
$prove \: that \: (\frac{x ^{a} }{x ^{b} }) {?}^{a ^{2} + ab + b ^{2} } \times ( \frac{x {}^{b} }{x {}^{c} } ) {} {?}^{b {}^{2} + bc + c {}^{2} } \times ( \frac{x {}^{c} }{x {}^{a} } ) {?}^{c {}^{2} + ca + a {}^{2} } = 1$
prove that

Answer 1

$({ \frac{ {x}^{a} }{ {x}^{b} } })^{ {a}^{2} + ab + {b}^{2} } \times ({ \frac{ {x}^{b} }{ {x}^{c} } })^{ {b}^{2} + bc + {c}^{2} } \times ({ \frac{ {x}^{c} }{ {x}^{a} } })^{ {c}^{2} + ac + {a}^{2} } = 1 \\ {x}^{(a - b)( {a}^{2} + ab + {b}^{2} )} \times {x}^{(b - c)( {b}^{2} + bc + {c}^{2}) } \times {x}^{(c - a)( {c}^{2} + ac + {a}^{2} )} = 1 \\ {x}^{( {a}^{3} - {b}^{3} ) } \times {x}^{ ({b}^{3} - {c}^{3} )} \times {x}^{ ({c}^{3} - {a}^{3}) } = 1 \: \: \: \: ( \: \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \: \: ) \\ {x}^{ ({a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3}) } = 1 \: \: \: \: ( \: \: {x}^{a} \times {x}^{b} = {x}^{a + b} \: \: \: ) \\ {x}^{0} = 1 \\ 1 = 1$