Question

Prove that:-

$\sf \red{\bigg( \frac{ {x}^{a^{2} } + {}^{ {b}^{2} } }{x {}^{ab} } \bigg) {}^{ {}^{a + b} } \sf \bigg( \frac{ {x}^{b^{2} } + {}^{ {c}^{2} } }{x {}^{bc} } \bigg) {}^{ {}^{b + c} } \sf \bigg( \frac{ {x}^{c^{2} } + {}^{ {a}^{2} } }{x {}^{ac} } \bigg) {}^{ {}^{a + c} } \sf = {x}^{2( {a}^{3} + {b}^{c} + {c}^{3} )} }$​​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\sf \red{\bigg( \frac{ {x}^{a^{2} } + {}^{ {b}^{2} } }{x {}^{ab} } \bigg) {}^{ {}^{a + b} } \sf \bigg( \frac{ {x}^{b^{2} } + {}^{ {c}^{2} } }{x {}^{bc} } \bigg) {}^{ {}^{b + c} } \sf \bigg( \frac{ {x}^{c^{2} } + {}^{ {a}^{2} } }{x {}^{ac} } \bigg) {}^{ {}^{a + c} } }$

Let consider

$\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ab} } \bigg]}^{a + b}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n} \: }}$

So, using this, we get

$\rm \: = \: {\bigg[{x}^{ {a}^{2} + {b}^{2} - ab} \bigg]}^{a + b}$

Now, we know that

$\rm :\longmapsto\:\boxed{\tt{ \: {( {x}^{m})}^{n} = {x}^{mn} \: }}$

So, using this, we get

$\rm \: = \: {x}^{( {a}^{2} + {b}^{2} - ab)(a + b)}$

$\rm \: = \: {x}^{ {a}^{3} + {b}^{3} }$

Now, Consider

$\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{bc} } \bigg]}^{b + c}$

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

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

$\rm \: = \: {x}^{ {b}^{3} + {c}^{3} }$

Now, Consider

$\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ca} } \bigg]}^{c + a}$

$\rm \: = \: {\bigg[{x}^{ {c}^{2} + {a}^{2} - ca} \bigg]}^{c + a}$

$\rm \: = \: {x}^{( {c}^{2} + {a}^{2} - ac)(c + a)}$

$\rm \: = \: {x}^{ {c}^{3} + {a}^{3} }$

Now, Consider

$\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ab} } \bigg]}^{a + b}{\bigg[\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{bc} } \bigg]}^{b + c}{\bigg[\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ca} } \bigg]}^{c + a}$

$\rm \: = \: {x}^{ {a}^{3} + {b}^{3} } \times {x}^{ {b}^{3} + {c}^{3} } \times {x}^{ {c}^{3} + {a}^{3} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ {x}^{m} \times {x}^{n} = {x}^{m + n} }}$

So, using this identity, we get

$\rm \: = \: {x}^{ {a}^{3} + {b}^{3} + {b}^{3} + {c}^{3} + {c}^{3} + {a}^{3} }$

$\rm \: = \: {x}^{ 2({a}^{3}+{b}^{3}+{c}^{3})}$

Hence,

$\boxed{\tt{ {\bigg[\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ab} } \bigg]}^{a + b}{\bigg[\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{bc} } \bigg]}^{b + c}{\bigg[\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ca} } \bigg]}^{c + a} = {x}^{ 2({a}^{3}+{b}^{3}+{c}^{3})}}}$