Math, asked by Anonymous, 1 year ago

(xa+b)2(xb+c)2(xc+a)2÷(xaxbxc)4=1

Answers

Answered by MaheswariS
66

\underline{\textsf{To prove:}}

\mathsf{\dfrac{(x^{a+b})^2(x^{b+c})^2(x^{c+a})^2}{(x^a\,x^b\,x^c)^4}=1}

\underline{\textsf{Solution:}}

\textsf{Consider,}

\mathsf{\dfrac{(x^{a+b})^2(x^{b+c})^2(x^{a+b})^2}{(x^a\,x^b\,x^c)^4}}

\mathsf{=\dfrac{x^{2(a+b)}\,x^{2(b+c)}\,x^{2(a+b)}}{(x^{a+b+c})^4}}

\mathsf{=\dfrac{x^{2a+2b}\,x^{2b+2c}\,x^{2c+2a}}{(x^{a+b+c})^4}}

\mathsf{=\dfrac{x^{2a+2b+2b+2c+2c+2a}}{x^{4(a+b+c)}}}

\mathsf{=\dfrac{x^{4a+4b+4c}}{x^{4a+4b+4c}}}

\mathsf{=1}

\implies\boxed{\mathsf{\dfrac{(x^{a+b})^2(x^{b+c})^2(x^{c+a})^2}{(x^a\,x^b\,x^c)^4}=1}}

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Answered by pulakmath007
52

SOLUTION

TO PROVE

 \sf{} {( {x}^{a + b} )}^{2}.  {( {x}^{b + c} )}^{2} . {( {x}^{b + c} )}^{2}  \div {( {x}^{a} . {x}^{b} . {x}^{c} )}^{4}  = 1

FORMULA TO BE IMPLEMENTED

We are aware of the formula of indices that

 \sf{}1. \:  \:  {a}^{m} . {a}^{n}  =  {a}^{m + n}

 \sf{}2. \:  \:  {( {a}^{m}) }^{n}  =  {a}^{mn}

 \displaystyle \sf{}3. \:  \:  \:  \frac{ {  a}^{m} }{ {a}^{n} } =  {a}^{m - n}

EVALUATION

 \sf{} {( {x}^{a + b} )}^{2}.  {( {x}^{b + c} )}^{2} . {( {x}^{b + c} )}^{2}  \div {( {x}^{a} . {x}^{b} . {x}^{c} )}^{4}

 =  \displaystyle \sf{}  \frac{  {( {x}^{a + b} )}^{2}.  {( {x}^{b + c} )}^{2} . {( {x}^{b + c} )}^{2}}{{( {x}^{a} . {x}^{b} . {x}^{c} )}^{4}  }

 =  \displaystyle \sf{}  \frac{  {( {x}^{2a + 2b} )}.  {( {x}^{2b + 2c} )}. {( {x}^{2b + 2c} )}}{{ {x}^{4a} . {x}^{4b} . {x}^{4c} }  }   \: ( \: by \: formula \: 2 \: )

 =  \displaystyle \sf{}   \frac{ {x}^{2a + 2b + 2b + 2c + 2c + 2a} }{ {x}^{4a + 4b + 4c} }  \:  \: (by \: formula \: 2)

 =  \displaystyle \sf{}   \frac{ {x}^{4a + 4b + 4c} }{ {x}^{4a + 4b + 4c} }

 \displaystyle \sf{} =  {x}^{(4a + 4b + 4c) - (4a + 4b + 4c)} \:  \: ( \: by \: formula \: 3)

 \displaystyle \sf{} =  {x}^{0}

 \sf{} = 1

Hence proved

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