1-3 -3- 1-2 -3- 1-4 -3- a19-64 b27-16 c64-19 d16-25

{ ( 1/3 )^ 3 ( 1/2 )^ 3}÷ ( 1/4 )^ 3=(3^3 2^3)÷ (4)^3 =(27 8)÷ 64=19/64 ...

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Question

${{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}} \div {(\frac{1}{4})}^{- 3} = ?$

$(a) \frac{19}{64}$

$(b) \frac{27}{16}$

$(c) \frac{64}{19}$

$(d) \frac{16}{25}$

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Solution

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{lim}_{n \to \infty} \frac{1 + 1 + 4 + 8 + 9 + 27 + 16 + 64 + . . .}{(1 + 2 + 3 + 4 + 5 + . . .)^{2}} =

{(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16}

\frac{27}{p^{3}}

\frac{128}{b^{3}}

(i) 2 \times 27^{\frac{1}{3}} \times 216^{- \frac{2}{3}}

(i i) 64^{- \frac{1}{3}} (64^{\frac{1}{3}} - 64^{- \frac{2}{3}})

(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{- 1}

{(\frac{64}{125})}^{\frac{- 2}{3}} \div \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} \times \frac{\sqrt{25}}{\sqrt[3]{64}}

{⎡ ⎢ ⎢ ⎣ 5 {⎛ ⎜ ⎝ 8^{\frac{1}{3}} + 27^{\frac{1}{3}} ⎞ ⎟ ⎠}^{3} ⎤ ⎥ ⎥ ⎦}^{\frac{1}{4}}

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