Solusi

Identitas (1)
$n^4-(n-1)^4=4n^3 - 6n^2 + 4n - 1$

Dari identitas di atas
$1^4-0^4=4.1^3 - 6.1^2 + 4.1 - 1$
$2^4-1^4=4.2^3 - 6.2^2 + 4.2 - 1$
$3^4-1^4=4.3^3 - 6.3^2 + 4.3 - 1$
...
$n^4-(n-1)^4=4n^3 - 6n^2 + 4n - 1$

Jumlahkan semua persamaan di atas, kita dapat persamaan (2)

$n^4=4(1^3+2^3+...n^3) - 6(1^2+2^2+...+n^2) + 4(1+2+...n) - n$

$n^4=4(\sum_{k=1}^{n} k^3) - 6(\sum_{k=1}^{n} k^2) + 4(\sum_{k=1}^{n} k) - n$

Rumus yang sudah kita tahu

$\sum_{k=1}^{n} k^2=\large{\frac{n(n+1)(2n+1)}{6}}$          (3)

$\sum_{k=1}^{n} k=\large{\frac{n(n+1)}{2}}$                       (4)

Substitusi rumus (3) dan (4) ke dalam persamaan (2)

$n^4=4(\sum_{k=1}^{n} k^3) - 6.\large{\frac{n(n+1)(2n+1)}{6}} + 4.\large{\frac{n(n+1)}{2}} - n$

$n^4=4(\sum_{k=1}^{n} k^3) - n(n+1)(2n+1) + 2n(n+1) - n$

$4(\sum_{k=1}^{n} k^3) = n^4 + n(n+1)(2n+1) - 2n(n+1) + n$

$4(\sum_{k=1}^{n} k^3) = n^4 + 2n^3 + n^2$

$4(\sum_{k=1}^{n} k^3) = n^2(n^2 + 2n + 1)$

$4(\sum_{k=1}^{n} k^3) = n^2(n+1)^2$

$\sum_{k=1}^{n} k^3 = \large{\frac{n^2(n+1)^2}{4}}$

$\sum_{k=1}^{n} k^3 = \large{(\frac{n(n+1)}{2})^2}$

$\sum_{k=1}^{n} k^3 = (\sum_{k=1}^{n} k)^2$