Find the value of $k$ , if $1^{3} + 2^{3} + 3^{3} + . . . . + k^{3} = 4356$

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Solution

Note that $k$ is a positive integer.
Given that $1^{3} + 2^{3} + 3^{3} + . . . . + k^{3} = 4356$
$\Rightarrow {(\frac{k (k + 1)}{2})}^{2} = 4356 = 6 \times 6 \times 11 \times 11$
Taking square root, we get $\frac{k (k + 1)}{2} = 66$
$\Rightarrow k^{2} + k - 132 = 0 \Rightarrow (k + 12) (k - 11) = 0$
Thus, $k = 11$ , since $k$ is positive.

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