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(23^(0)+24^(0)+25^(0))^(-2)÷9

(230+240+250)−2÷9 \left(23^{0}+24^{0}+25^{0}\right)^{-2} \div 9

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Full solution

Q. (230+240+250)−2÷9 \left(23^{0}+24^{0}+25^{0}\right)^{-2} \div 9
  1. Evaluate Exponents with Base 00: Evaluate the exponents with base 00.\newlineAccording to the exponent rule, any non-zero number raised to the power of 00 is 11.\newlineSo, 230=123^{0} = 1, 240=124^{0} = 1, and 250=125^{0} = 1.
  2. Add Results from Step 11: Add the results from Step 11.\newline1+1+1=31 + 1 + 1 = 3
  3. Raise Sum to Power of −2-2: Raise the sum to the power of −2-2. (3)−2(3)^{-2} means 11 divided by (3)2(3)^2.
  4. Calculate (3)2(3)^2: Calculate (3)2(3)^2.(3)2=3×3=9(3)^2 = 3 \times 3 = 9
  5. Calculate 11 divided by (3)2(3)^2: Calculate 11 divided by (3)2(3)^2.19=19\frac{1}{9} = \frac{1}{9}
  6. Divide Result by 99: Divide the result from Step 55 by 99.\newline(1/9)÷9=(1/9)×(1/9)=1/81(1/9) \div 9 = (1/9) \times (1/9) = 1/81

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