$6^{3} + 5^{3} + 4^{3} + 3^{3} + 2^{3} =$ ____________.

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Solution

Sum of cubes of ‘n’ natural numbers = ${\frac{n (n + 1)}{2}}^{2}$

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Similar questions

(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . . . t o 10

(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . . . .

10

1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + . . . . . . . . + n^{3}

(5^{3} - 3^{3})

(4^{3} - 2^{3})

1^{3} + 2^{3} + \frac{3^{3}}{⌊ 2} + \frac{4^{3}}{⌊ 3} + \frac{5^{3}}{⌊ 4} + . . . .

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