Show that: ( a^ x+1 a^ y+1 )^ x+y ( a^ y+2 a^ z+2 )^ y+z ( a^ z+3 a^ x+3 )^ z+x =1

LHS= ( a^ x+1 a^ y+1 )^ x+y ( a^ y+2 a^ z+2 )^ y+z ( a^ z+3 a^ x+3 )^ z+x = (a^ x+1-(y+1) )^ x+y (a^ y+2-(x+2) )^ y+z (a^ z+3-(x+3) )^ z+x = (a^ x+1-y-1 )^ x+y (a^ y+2-z-2 )^ y+z (a^ z+3-x-3 )^ z+x = ( a^ x-y )^ x+y (a^ y-z )^ y+z (a^ z-x )^ z+x = (a^ (x-y)(x+y) ) (a^ (y-z)(y+z) ) (a^ (z-x)(z+x) ) = (a^ x^2-y^2 ) (a^ y^2-z^2 ) (a^ (z^2-x^2) ) =a^ x^2-y^2+y^2-z^2+z^2-x^2 =a^0 =1 =RHS

Show that: ( a^ x+1 a^ y+1 )^ x+y ( a^ y+2 a^ z+2 )^ y+z ( a^ z+3…

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Question

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

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Answer

$\text{LHS}=\Big(\frac{\text{a}^{\text{x}+1}}{\text{a}^{\text{y}+1}}\Big)^{\text{x}+\text{y}}\Big(\frac{\text{a}^{\text{y}+2}}{\text{a}^{\text{z}+2}}\Big)^{\text{y}+\text{z}}\Big(\frac{\text{a}^{\text{z}+3}}{\text{a}^{\text{x}+3}}\Big)^{\text{z}+\text{x}}$$=\big(\text{a}^{\text{x}+1-(\text{y}+1)}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}+2-(\text{x}+2)}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}+3-(\text{x}+3)}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{\text{x}+1-\text{y}-1}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}+2-\text{z}-2}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}+3-\text{x}-3}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{\text{x}-\text{y}}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}-\text{z}}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}-\text{x}}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{(\text{x}-\text{y})(\text{x}+\text{y})}\big)\big(\text{a}^{(\text{y}-\text{z})(\text{y}+\text{z})}\big)\big(\text{a}^{(\text{z}-\text{x})(\text{z}+\text{x})}\big)$
$=\big(\text{a}^{\text{x}^2-\text{y}^2}\big)\big(\text{a}^{\text{y}^2-\text{z}^2}\big)\big(\text{a}^{(\text{z}^2-\text{x}^2)}\big)$
$=\text{a}^{\text{x}^2-\text{y}^2+\text{y}^2-\text{z}^2+\text{z}^2-\text{x}^2}$
$=\text{a}^0$
$=1$
$=\text{RHS}$

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