Question

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

Answer 1

Hey here is your answer

$\color{blue} {\mathbb{\underline{\star{NICE \: QUESTION}\star}}}$

To prove :-

${ (\frac{ {x}^{a} }{ {x}^{b} } )}^{a + b - c} \times { (\frac{ {x}^{b} }{ {x}^{c} } )}^{b + c - a} \times { (\frac{ {x}^{c} }{ {x}^{a} } )}^{c + a - b} = 1$

so, let's proof it :-

LHS =

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

using identity :-

${a}^{m} \times {a}^{n} = {a}^{(m + n)}$

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

using identity :-

$\frac{ {a}^{m} }{ {a}^{n } } = {a}^{(m - n)}$
$\frac{ {x}^{({a}^{2}+ {b}^{2}+ {c}^{2} )} }{ {x}^{({a}^{2} + {b}^{2} + {c}^{2} ) } } \\ \\ {x}^{({a}^{2}+ {b}^{2}+ {c}^{2} ) - ({a}^{2}+ {b}^{2}+ {c}^{2} )} \\ \\ {x}^{({a}^{2}+ {b}^{2}+ {c}^{2} - {a}^{2} - {b}^{2} - {c}^{2} )} \\ \\ {x}^{0}$

As we all know if any number have 0 as it exponent the answer will be 1.

${x}^{0} = 1$

LHS = 1
RHS = 1

LHS = RHS
1 = 1

So,

$\bf { (\frac{ {x}^{a} }{ {x}^{b} } )}^{a + b - c} \times { (\frac{ {x}^{b} }{ {x}^{c} } )}^{b + c - a} \times { (\frac{ {x}^{c} }{ {x}^{a} } )}^{c + a - b} = 1$

hence proved !!!!!!!

hope this helps you dear

### BE BRAINLY ###

Answer 2

Answer:

i couldn't type powers properly and my root system is hangout...so...

in the image i solved the answer.

Step-by-step explanation:

Hope u understood....

if not then u can question it in one of answers...

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