Question

simplify (xb/xc)b+c-a x (xc/xa)c+a-b x (xa xb)a+b-c


pls tell​

Answer 1

Appropriate Question:

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

Answer:

$\boxed{\bf \: {\bigg( \dfrac{ {x}^{b} }{ {x}^{c} } \bigg) }^{b + c - a} \times {\bigg( \dfrac{ {x}^{c} }{ {x}^{a} } \bigg) }^{c + a - b} \times {\bigg( \dfrac{ {x}^{a} }{ {x}^{b} } \bigg) }^{a + b - c} = 1 \: }$

Step-by-step explanation:

Given expression is

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

We know,

$\boxed{\sf \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }$

So, using this law of exponents, we get

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

We know,

$\boxed{\sf \: {( {x}^{m} )}^{n} \: = \: {x}^{mn} \: }$

So, using this law of exponents, we get

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

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

$\sf \: = \: {\bigg(x \bigg) }^{ {b}^{2} - {c}^{2} - ab + ac + {c}^{2} - {a}^{2} - bc + ab + {a}^{2} - {b}^{2} - ac + bc}$

$\sf \: = \: {x}^{0}$

$\sf \: = \: 1$

Hence,

$\implies\boxed{\bf \: {\bigg( \dfrac{ {x}^{b} }{ {x}^{c} } \bigg) }^{b + c - a} \times {\bigg( \dfrac{ {x}^{c} }{ {x}^{a} } \bigg) }^{c + a - b} \times {\bigg( \dfrac{ {x}^{a} }{ {x}^{b} } \bigg) }^{a + b - c} = 1 \: }$