If m3 - 23 - 33 - 5-2 - 6-5 - then m can be equal to -

The correct option is B 1 (m^3+2^3+3^3)^ 5/2 = (6)^ 5 #160;= (m^3+2^3+3^3)^1/2 = 6 #160; #160;= m^3+2^3+3^3 = 36 #160; #160;= ...

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Multiplication of Monomials

If m3 + 23 ...

Question

If $(m^{3} + 2^{3} + 3^{3})^{- \frac{5}{2}} = (6)^{- 5}$ , then $m$ can be equal to :

- 1

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\frac{1}{2}

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2

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1

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Solution

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(m^{3} + 2^{3} + 3^{3})^{- \frac{5}{2}} = (6)^{- 5}

S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + . . . . . . + \frac{1 + 2 + . . . . . + n}{1^{3} + 2^{3} + . . . . + n^{3}}

100 S_{n} = n

n

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

1^{3} + 2^{3} + 3^{3} + . . . . . . . . . + n^{3}

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