Question

${ \boxed{ \huge{ \frak{ \red{question}}}}}$
please solve the question given in the attachment !!​

Answer 1

$\large\underline{\sf{Solution-}}$

Identities Used :-

$\red{ \boxed{ \sf{ \: log( {x}^{y} ) = y log(x) }}}$

$\red{ \boxed{ \sf{ \: log(xy) = log(x) + log(y) }}}$

$\red{ \boxed{ \sf{ log(1) = 0 \: }}}$

$\red{ \boxed{ \sf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

$\red{ \boxed{ \sf{ \: (x - y)( {x}^{2} + xy + {y}^{2}) = {x}^{3} - {y}^{3}}}}$

Let's solve the problem now!!!

Given that

$\rm :\longmapsto\:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b}$

Let assume that

$\rm :\longmapsto\:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b} = k$

$\rm :\longmapsto\:logx = k(b - c) - - (1)$

$\rm :\longmapsto\:logy = k(c - a) - - (2)$

$\rm :\longmapsto\:logz = k(a - b) - - (3)$

Now,

Multiply equation (1) by a, we get

$\rm :\longmapsto\:alogx = ak(b - c)$

$\rm :\longmapsto\:log {x}^{a} = k(ab - ac) - - - (4)$

Multiply equation (2) by b, we get

$\rm :\longmapsto\:blogy = kb(c - a)$

$\rm :\longmapsto\:log {y}^{b} = k(bc - ab) - - - (5)$

Multiply equation (3) by c, we get

$\rm :\longmapsto\:clogz = kc(a - b)$

$\rm :\longmapsto\:log {z}^{c} = k(ac - bc) - - - (6)$

On adding equation (4), (5) and (6), we get

$\rm :\longmapsto\:log {x}^{a} + log {y}^{b} + log {z}^{c} = k(ab - ac + bc - ba + ac - bc)$

$\rm :\longmapsto\:log( {x}^{a}{y}^{b}{z}^{c} ) = 0$

$\rm :\longmapsto\:log( {x}^{a}{y}^{b}{z}^{c} ) = log1$

$\bf :\longmapsto\:{x}^{a}{y}^{b}{z}^{c} = 1$

Hence Proved (a)

Multiply equation (1) by (b + c), we get

$\rm :\longmapsto\:(b + c)logx = (b + c)k(b - c)$

$\rm :\longmapsto\:log {x}^{b + c} = k( {b}^{2} - {c}^{2} ) - - - (7)$

Multiply equation (2) by (c + a), we get

$\rm :\longmapsto\:(c + a)logy = k(c + a)(c - a)$

$\rm :\longmapsto\:log {y}^{c + a} = k( {c}^{2} - {a}^{2} ) - - - (8)$

Multiply equation (3) by a + b, we get

$\rm :\longmapsto\:(a + b)logz = k(a + b)(a - b)$

$\rm :\longmapsto\:log {z}^{a + b} = k( {a}^{2} - {b}^{2} ) - - - (9)$

On adding equation (7), (8) and (9), we get

$\rm :\longmapsto\:log {x}^{b + c} + log {y}^{c + a} + log {z}^{a + b} = k(0)$

$\rm :\longmapsto\:log ({x}^{b + c} {y}^{c + a}{z}^{a + b} ) = 0$

$\rm :\longmapsto\:log ({x}^{b + c} {y}^{c + a}{z}^{a + b} ) = log1$

$\bf :\longmapsto\: {x}^{b + c} {y}^{c + a}{z}^{a + b} =1$

Hence, Proved (b)

Multiply equation (1) by b^2 + bc + c^2, we get

$\rm :\longmapsto\:( {b}^{2} + bc + {c}^{2} )logx = ( {b}^{2} + bc + {c}^{2} )k(b - c)$

$\rm :\longmapsto\:log {x}^{ {b}^{2} + bc + {c}^{2} } = k( {b}^{3} - {c}^{3} ) - - - (10)$

Similarly,

$\rm :\longmapsto\:log {y}^{ {c}^{2} + ac + {a}^{2} } = k( {c}^{3} - {a}^{3} ) - - - (11)$

Similarly,

$\rm :\longmapsto\:log {z}^{ {a}^{2} + ab + {b}^{2} } = k( {a}^{3} - {b}^{3} ) - - - (12)$

On adding equation (10), (11) and (12), we get

$\rm :\longmapsto\:log {x}^{ {b}^{2} + bc + {c}^{2} } + log {y}^{ {c}^{2} + ac + {a}^{2}} + log {z}^{ {a}^{2} + ab + {b}^{2} } = k(0)$

$\rm :\longmapsto\:log ({x}^{ {b}^{2} + bc + {c}^{2} }{y}^{ {c}^{2} + ac + {a}^{2}}{z}^{ {a}^{2} + ab + {b}^{2} }) = 0$

$\rm :\longmapsto\:log ({x}^{ {b}^{2} + bc + {c}^{2} }{y}^{ {c}^{2} + ac + {a}^{2}}{z}^{ {a}^{2} + ab + {b}^{2} }) = log1$

$\bf :\longmapsto\:{x}^{ {b}^{2} + bc + {c}^{2} }{y}^{ {c}^{2} + ac + {a}^{2}}{z}^{ {a}^{2} + ab + {b}^{2} } = 1$

Hence, Proved (c)