Find: $1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . . . + 9^{3} = ?$

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Solution

$1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . . . . . . + 9^{3} = ?$

$(1^{3} + 2^{3} + 3^{3} + . . . . . . . . + 9^{3}) - 2 (2^{3} + 4^{3} + 6^{3} + 8^{3})$

$(1^{3} + 2^{3} + 3^{3} + . . . . . . . . + 9^{3}) - {2.2}^{3} (1^{3} + 2^{3} + 3^{3} + 4^{3})$

$n \sum r = 1 n^{3} = {(\frac{n (n + 1)}{2})}^{2}$
${(\frac{(9) (9 + 1)}{2})}^{2} - 16 {(\frac{(4) (4 + 1)}{2})}^{2}$
$(2025 - 1600) = 425$ Ans

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1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . . + 9^{3} =

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