Question

\(\frac{27^4\cdot4^3}{9^5.8^2}\)

 

Answer 1

\(\frac{27^4.4^3}{9^5.8^2}=\frac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\frac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)

Answer 2

=\(\frac{3^{12}.2^6}{3^{15}.2^6}\)

=\(\frac{3^{12}}{3^{12}.3^3}\)

=\(\frac{1}{27}\)