(5⁰ + 6⁻¹) × 3²To solve this we must apply laws of exponents. First we will apply these TWO laws of exponents for the brackets.a⁰ = 1a^{-m} = \dfrac{1}{a^{m}}a −m = a m 1 ⇒ (5⁰ + 6⁻¹) × 3²⇒ (1+\dfrac{1}{6})\TIMES 3^{2}(1+ 61 )×3 2 Now we will find the LCM of the denominators in the fractions in the bracket. The LCM of 1 and 6 is 6 so using this value we will make the denominators equal.⇒ (1+\dfrac{1}{6})\times 3^{2}(1+ 61 )×3 2 ⇒ (\dfrac{1\times 6 }{1\times 6 } +\dfrac{1}{6})\times 3^{2}( 1×61×6 + 61 )×3 2 ⇒ (\dfrac{6 }{6 } +\dfrac{1}{6})\times 3^{2}( 66 + 61 )×3 2 Now add the denominators to OBTAIN the answer of the brackets.⇒ (\dfrac{6 }{6 } +\dfrac{1}{6})\times 3^{2}( 66 + 61 )×3 2 ⇒ \frac{7 }{6 }\times 3^{2} 67 ×3 2 Now we'll find the value of 3².3² = 3×3 = 9⇒ \dfrac{7 }{6 }\times 9 67 ×9⇒ \dfrac{7 }{6 \div 3 }\times 9\div 3 6÷37 ×9÷3⇒ \dfrac{7 }{2}\times 3 27 ×3⇒ \dfrac{21}{2} 221 \bf \therefore (5^{0} + 6^{-1}) \times 3^{2} = \frac{21}{2}∴(5 0 +6 −1 )×3 2 = 221 _____________________________________