Question

If ${\left(\frac{b^3{c}^{-2}}{b^{-4}{c}^3}\right)}^{-3}÷{\left(\frac{b^{-1}c}{b^2{c}^{-2}}\right)}^5=b^x{c}^y,$ find $x,y$

Answer 1

We have,

${\left(\frac{b^3{c}^{-2}}{b^{-4}{c}^3}\right)}^{-3}÷{\left(\frac{b^{-1}c}{b^2{c}^{-2}}\right)}^5=b^x{b}^y$

${\left(\frac{b^7}{c^5}\right)}^{-3}÷{\left(\frac{c^3}{b^3}\right)}^5=b^x{b}^y$

$\left(\frac{c^{15}}{b^{21}}\right)÷\left(\frac{c^{15}}{b^{15}}\right)=b^x{b}^y$

$\left(\frac{c^{15}}{b^{21}}\right)\times \left(\frac{b^{15}}{c^{15}}\right)=b^x{b}^y$

$b^{-6}{c}^0=b^x{b}^y$

Therefore,

$x=-6,y=0$

Hence, this is the answer.