Question

(3²)³+(2/3)⁰+(3)⁵×(1/3)⁴​

Answer 1

Answer:

To find:

The value of (3²)³+(2/3)⁰+(3)⁵×(1/3)⁴​

Proof:

= (3²)³+(2/3)⁰+(3)⁵×(1/3)⁴​

=(3²)³ $+(\frac{2}{3})^{0}$ + (3)⁵ + $(\frac{1}{3})^{4}$      

= $3^{2 X 3}$ + 1 + (3)⁵ + $(3^{-1} )^{4}$                      [ $(a^{m} )^{n} = a^{m Xn}$],  [ $(a)^0= 1$ ],  [$\frac{1}{a^{m} } = a^{-m}$]

= $3^{6}$ + 1+ (3)⁵ + $(3)^{-4}$

= $3^{6}$ + $(3)^{5+(-4)}$ + 1                                [ $a^{m} X a^ n} = a^{m+n}$]

= $3^{6}$ + 3 + 1

= $3^{6}$ + 4

= 729 + 4

= 733

Hence, the value of  (3²)³ $+(\frac{2}{3})^{0}$ + (3)⁵ + $(\frac{1}{3})^{4}$      

= 733

Hope you got that.

Thank You.