Show that:$ \left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}\left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}=1 $

To do: 

We have to show that \( \left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}\left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}=1 \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

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

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=(\frac{a^{x+1}}{a^{y+1}})^{x+y}(\frac{a^{y+2}}{a^{z+2}})^{y+z}(\frac{a^{z+3}}{a^{x+3}})^{z+x}$

$=(a^{x+1-y-1})^{x+y} \times(a^{y+2-z-2})^{y+z}(a^{z+3-x-3})^{z+x}$

$=(a^{x-y})^{x+y} \times(a^{y-z})^{y+z} \times(a^{z-x})^{z+x}$

$=a^{(x-y)(x+y)} \times a^{(y-z)(y+z)} \times d^{(z-x)(z+x)}$

$=a^{x^{2}-y^{2}} \times a^{y^{2}-z^{2}} \times a^{z^{2}-x^{2}}$

$=a^{x^{2}-y^{2}+y^{2}-z^{2}+z^{2}-x^{2}}$

$=a^{0}$

$=1$

$=$ RHS

Hence proved.         

Tutorialspoint

Updated on: 10-Oct-2022

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