Question

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

Answer 1

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

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

$=\left[a^{(x+1-y-1)(x+y)}\right]\left[a^{(y+2-z-2)(y+z)}\right]\left[a^{(z+3-x-3)(z+x)}\right]$

$=\left[a^{(x-y)(x+y)}\right]\left[a^{(y-z)(y+z)}\right]\left[a^{(z-x)(z+x)}\right]$

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

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

$=a^0$
$=1$
$=\text{RHS}$
Hence, proved .