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Answer 1

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Answer 2

Question :-

$\sf \left[\left(\dfrac{1}{3}\right)^{-3} - \left(\dfrac{1}{2}\right)^{-3}\right] \div \left(\dfrac{1}{4}\right)^{-2}$

Answer :-

$\implies\sf\dfrac{\left(\dfrac{1}{3}\right)^{-3} - \left(\dfrac{1}{2}\right)^{-3}}{\left(\dfrac{1}{4}\right)^{-2}}$

$\implies\sf\dfrac{\dfrac{1^{-3}}{3^{-3}} - \dfrac{1^{-3}}{2^{-3}}}{\dfrac{1^{-2}}{4^{-2}} }$

$\implies\sf\dfrac{\dfrac{3^3}{1} - \dfrac{2^3}{1}}{\dfrac{4^2}{1}}$

$\implies\sf\dfrac{3^3-2^3}{4^3}$

• 3³ = 3 × 3 × 3 = 27
• 2³ = 2 × 2 × 2 = 8
• 4² = 4 × 4 = 16

$\implies\sf\dfrac{27-8}{16}$

$\implies\sf\dfrac{19}{16}$

$\boxed{\sf \left[\left(\dfrac{1}{3}\right)^{-3} - \left(\dfrac{1}{2}\right)^{-3}\right] \div \left(\dfrac{1}{4}\right)^{-2} =\dfrac{19}{16}}$

Additional information :-

$\bf Law \: of \: Exponents :\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}$