Question

prove that (x^a/x^b)^a+b-c(x^b/x^c)^b+c-a(x^c/x^a)^c+a-b =1​.
pls ans fast

Answer 1

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

Consider,

$\red{\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{a} }{ {x}^{b} } \bigg]}^{a + b - c}}$

We know,

$\red{ \boxed{ \sf{ \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n}}}}$

So, using this identity, we get

$\rm \:  =  \: {\bigg[ {x}^{a - b} \bigg]}^{a + b - c}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(a - b)(a + b - c)}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(a - b)(a + b) - (a - b)c}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ {a}^{2} - {b}^{2} - (a - b)c}$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

Now, Consider

$\red{\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{b} }{ {x}^{c} } \bigg]}^{b + c - a}}$

$\rm \:  =  \: {\bigg[ {x}^{b - c} \bigg]}^{b + c- a}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(b - c)(b + c - a)}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(b - c)(b + c)- (b - c)a}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ {b}^{2} - {c}^{2} - (b - c)a}$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

Now, Consider

$\red{\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{c} }{ {x}^{a} } \bigg]}^{c + a - b}}$

$\rm \:  =  \: {\bigg[ {x}^{c - a} \bigg]}^{c + a- b}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(c - a)(c + a - b)}$

$\rm \:  =  \: {\bigg[x\bigg]}^{(c - a)(c + a)- (c - a)b}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ {c}^{2} - {a}^{2} - (c - a)b}$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

Now, Consider

$\red{\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{a} }{ {x}^{b} } \bigg]}^{a + b - c}{\bigg[\dfrac{ {x}^{b} }{ {x}^{c} } \bigg]}^{b + c - a}{\bigg[\dfrac{ {x}^{c} }{ {x}^{a} } \bigg]}^{c + a - b}}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ {a}^{2} - {b}^{2} - (a - b)c} \times {\bigg[x\bigg]}^{ {b}^{2} - {c}^{2} - (b - c)a} \times {\bigg[x\bigg]}^{ {c}^{2} - {a}^{2} - (c - a)b}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ {a}^{2} - {b}^{2} - ac + bc + {b}^{2} - {c}^{2} - ba + ac + {c}^{2} - {a}^{2} - bc + ba}$

$\rm \:  =  \: {\bigg[x\bigg]}^{ 0}$

$\rm \:  =  \: 1$

Hence,

$\red{\bf :\longmapsto\: {\bigg[\dfrac{ {x}^{a} }{ {x}^{b} } \bigg]}^{a + b - c}{\bigg[\dfrac{ {x}^{b} }{ {x}^{c} } \bigg]}^{b + c - a}{\bigg[\dfrac{ {x}^{c} }{ {x}^{a} } \bigg]}^{c + a - b} = 1}$