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Properties of Cubes and Cube Roots

13+23+… . .+8...

Question

$1^{3} + 2^{3} + . . . . . . + 8^{3} = ?$

36^{2}

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10^{2}

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63^{2}

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72^{2}

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Solution

The correct option is A $36^{2}$
The sum of the cubes of first n natural numbers is equal to the square of their sum.

$1^{3} + 2^{3} + 3^{3} . . . . . . + n^{3} = (1 + 2 + 3...... + n)^{2} 1^{3} + 2^{3} + 3^{3} . . . . . . + 8^{3} = (1 + 2 + 3...... + 8)^{2} = 36^{2}$

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a) \frac{5}{13} + \frac{4}{13} =_____

b) \frac{15}{23} - \frac{11}{23} =_____

c) \frac{3}{4} + \frac{3}{8} =_____

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g) 9 \frac{1}{3} - 7 \frac{8}{9} =_____

h) 9 \frac{1}{2} - 8 \frac{3}{4} =_____

g) 3 \frac{3}{8} - 2 \frac{3}{4} =_____

$2^{3} + 4^{3} + 6^{3} + 8^{3} + . . . .$

f (x) = \{\begin{matrix} \frac{1 - \sin^{2} x}{3 \cos^{2} x}, & x < \frac{π}{2} \\ a, & x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

a = \frac{1}{3},

a = \frac{1}{3}, b = \frac{8}{3}

a = \frac{2}{3}, b = \frac{8}{3}

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