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Question Number 10142 by 0942679167 last updated on 26/Jan/17

Commented by prakash jain last updated on 27/Jan/17

$\underset{i = x}{\sum^{x + 7}} i^{3} = \underset{i = 1}{\sum^{x + 7}} i^{3} - \underset{i = 1}{\sum^{x - 1}} i^{3}$ $= {(\frac{(x + 7) (x + 8)}{2})}^{2} - {(\frac{(x - 1) x}{2})}^{2}$ $= \frac{1}{4} [(x + 7) (x + 8) - x (x - 1)] [(x + 7) (x + 8) + (x - 1) x]$ $= \frac{1}{4} [16 x + 56] [2 x^{2} + 14 x + 56]$ $= \frac{1}{4} 8 (2 x + 7) 2 (x^{2} + 7 x + 28)$ $= 4 (2 x + 7) (x^{2} + 7 x + 28)$ $y^{3} = 4 (2 x + 7) (x^{2} + 7 x + 28)$ $y = \sqrt[3]{4 (2 x + 7) (x^{2} + 7 x + 28)}$
Commented by prakash jain last updated on 27/Jan/17

$y^{3} = 4 (2 x + 7) (x^{2} + 7 x + 28) \Rightarrow y even = 2 k$ $8 k^{3} = 8 x^{3} + 84 x^{2} + 420 x + 784$ $8 k^{3} - 8 x^{3} = 28 (3 x^{2} + 15 x + 28)$ $2 (k^{3} - x^{3}) = 7 (3 x^{2} + 15 x + 28)$ $k^{3} - x^{3} = 7 r$ $2 r = 3 x^{2} + 15 x + 28$ $3 x^{2} + 15 x + 28 - 2 r = 0$ $x = \frac{- 15 \pm \sqrt{225 - 4 \times 3 (28 - 2 r)}}{6}$ $x = \frac{- 15 \pm \sqrt{225 - 336 + 24 r}}{6}$ $= \frac{- 15 \pm \sqrt{24 r - 111}}{6}$ $24 r - 111 = 9 m^{2}$ $r = \frac{9 m^{2} + 111}{24} = \frac{3 m^{2} + 37}{8}$ $m = 1, x = - 3, y = 4$ $m = 1, x = - 2, y = 6$ $m = 3, x = - 4, y = 4$ $m = 3, x = - 1, n o y$ $m = 5, x = - 5, y = - 6$
Commented by mrW1 last updated on 27/Jan/17

$g r e a t!$
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