Question

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2 nd question​

Answer 1

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Answer 2

$\red{ \bold{ \huge{ \underline{ \overline{ \mid{ \purple{ \: \: SOLUTION\: \: \red{ \mid}}}}}}}}$

$\rm{L.H.S. =( x {}^{ \frac{1}{a - b} }) {}^{ \frac{1}{a - c} } \: . \:(x {}^{ \frac{1}{b - c} }) {}^{ \frac{1}{b - a} } \: . \:(x {}^{ \frac{1}{c - a} }) {}^{ \frac{1}{c- b} } \: } \\ \\ \rm{ =x {}^{ \frac{1}{(a - b)(a - c)} } \: . \: x {}^{ \frac{1}{(b - c)(b - a)} } \: . \: x {}^{ \frac{1}{(c - a)(c - b)} } } \\ \\ \rm{ = x {}^{ \frac{1}{(a - b)[ - (c - a)]} } \: . \: x {}^{ \frac{1}{(b - c)[ - (a - b)]}}\: . \:x {}^{ \frac{1}{(c - a)[ - (b - c)]} } } \\ \\ \rm{ =x {}^{ \frac{1}{ [- (a - b)(c - a)] } \: + \: \frac{1}{[ - (b - c)(a - b)]} \: + \: \frac{1}{[ - (c - a)(b - c)]} } } \\ \\ \rm{ =x {}^{ - \: \frac{1}{(a - b)(c - a)} \: - \: \frac{1}{(b - c)(a - b)} \: - \: \frac{1}{(c - a)(b - c)} } } \\ \\ \rm{ = x {}^{ - [\frac{1}{(a - b)(c - a)} + \frac{1}{(b - c)(a - b)} + \frac{1}{(c - a)(b - c)} ] } } \\ \\ \rm{ = x {}^{ - [ \frac{(b - c) + (c - a) + (a - b)}{(a - b)(b - c)(c - a)} ]} } \\ \\ \rm{ = x {}^{ - [\frac{ \: \: \: b - c + c - a + a - b \: \: }{(a - b)(b - c)(c - a)} ]} } \\ \\ \rm{ = x {}^{ - \: [\frac{ \: 0 \: }{(a - b)(b - c)(c - a)}]} } \\ \\ \rm{ = x {}^{0} } \\ \\ \rm{ = 1} \\ \\ \rm{=R.H.S. }$

$\red{ \bold{ \huge{ \underline{ \overline{ \mid{ \purple{ \: \: FORMULA \: \: USED\: \: \red{ \mid}}}}}}}}$

$\rm{1. \: \: \: \: \red{a {}^{m} \times a {}^{n} = a {}^{m + n} }} \\ \\ \rm{2. \: \: \: \: \red{(a {}^{m} ) {}^{n} = a {}^{mn} }} \\ \\ \rm{3. \: \: \: \: \red{a {}^{0} = 1}}$