Question

Prove the following ​

Answer 1

$(\frac{x {}^{b} }{x {}^{a} } ) {}^{b + a - c} \times ( \frac{x {}^{c} }{x {}^{b} } ) {}^{c + b - a} \times ( \frac{x {}^{a} }{ {x}^{c} }) {}^{c + a - b}$

$\frac{x {}^{b {}^{2} + ba - bc} }{x {}^{a b + a {}^{2} - ac} } \times \frac{x {}^{c {}^{2} + bc - ac} }{x {}^{bc + b {}^{2} - ab} } \times \frac{x {}^{ac + a {}^{2} - ab } }{x {}^{c {}^{2} + ac - bc} }$

$using \: exponent \: laws \\ \frac{x {}^{y} }{x {}^{b} } = x {}^{y - b}$

$x {}^{b {}^{2} + ab - ba - (ab + a {}^{2} - ca) } \times x {}^{c {}^{2} + bc - ab - (bc + b {}^{2} - ab) } \times x {}^{ac + a {}^{2} - ab - (c {}^{2} + ac - bc) }$

$x {}^{b {}^{2} + ab - bc - ab - a {}^{2} + ac } \times x {}^{c {}^{2} + bc - ac - bc - b {}^{2} + ab } \times x {}^{ac + a {}^{2} - ab - c {}^{2} - ac + ba }$

$x {}^{b {}^{2} - a {}^{2} + ac - bc} \times x {}^{c {}^{2} - b {}^{2} + ab - ac} \times x {}^{a {}^{2} - c {}^{2} + bc - ab }$

$using \: exponential \: law \\ x {}^{y} \times x {}^{b} = x {}^{y + b}$

$x {}^{b {}^{2} - a {}^{2} + ac - bc + c {}^{2} - b {}^{2} + ab - ac + a {}^{2} - c {}^{2} + bc - ab }$

$x {}^{0} = 1$

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