Question

(x^a+b/x^c)^a-b *(x^b+c/x^a)^b-c *(x^c+a/x^b)^c-a =1​

Answer 1

$\bold{\pink{To \: prove}}\longrightarrow \\ ( { \dfrac{ {x}^{a + b} }{ {x}^{c} } })^{a - b} \: ({ \dfrac{ {x}^{b + c} }{ {x}^{a} } })^{b - c} \: ( { \dfrac{ {x}^{c + a} }{ {x}^{b} } })^{c - a} \: = 1 \\\bold{\green {Concept \: used }}\longrightarrow\\ 1) {( \dfrac{p}{q}) }^{m} = \dfrac{ {p}^{m} }{ {q}^{m} } \\ 2)(x + y) \: (x - y) = {x}^{2} - {y}^{2} \\ 3) {p}^{m} {p}^{n} = {p}^{m + n} \\ 4) {a}^{0} = 1$

$\bold{\blue{Solution }}\longrightarrow\\ \pink{ LHS} \\ = {( \dfrac{ {x}^{a + b} }{ {x}^{c} } )}^{a - b} \: {( \dfrac{ {x}^{c + a} }{ {x}^{b} }) }^{b - c} \: { (\dfrac{ {x}^{c + a} }{ {x}^{b} }) }^{c - a}$

$= \dfrac{ {x}^{(a + b)(a - b)} }{ {x}^{c(a - b)} } \: \times \: \dfrac{ {x}^{(b + c)(b - c)} }{ {x}^{a(b - c)} } \: \times \: \dfrac{ {x}^{(c + a)(c - a)} }{ {x}^{b(c - a)} }$

$= \dfrac{ {x}^{ {a}^{2} - {b}^{2} } }{ {x}^{ca - bc} } \times \dfrac{ {x}^{ {b}^{2} - {c}^{2} } }{ {x}^{ab - ca} } \times \dfrac{ {x}^{ {c}^{2} - {a}^{2} } }{ {x}^{bc - ab} } \\ = \dfrac{ {x}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } }{ {x}^{ca - bc + ab - ca + bc - ab} } \\ = \dfrac{ {x}^{0} }{ {x}^{0} } \\ = \dfrac{1}{1} \\ = 1 \\ = \pink{RHS}$

$\bold{\red{Additioal \: informtion}}\longrightarrow \\ 1) \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} \\ 2) {(ab)}^{m} = {a}^{m} {b}^{m} \\ 3) \sqrt[m]{a} = {a}^{ \frac{1}{m} } \\ 4) { ({x}^{m} )}^{n} = {x}^{mn}$