Factorise it!

• $331$ is a prime number. Therefore it factors out into $1 \times 331$ .
• Also $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ .
• If we equate the two, $x-y = 1$ , $x^2 + xy + y^2 = 331$ since $x^2 + xy + y^2 > x-y$

$\begin{cases} x - y= 1 \implies {\color{#3D99F6}{x = y + 1}} \\x^2 + xy + y^2 = 331 \end{cases}$

\[\begin{align} \\ 331 &= ({\color{Blue}{y + 1}})^2 + ({\color{Blue}{y + 1}})(y) + y^2 \\ 331 &= 3y^2 + 3y + 1 \\ &= y^2 + y -110 \\ &= (y+11)(y-10) \\ &\implies y = 10 \\ &\implies \boxed{x = 11}

\end{align} \]

$x^{3} - y^{3}$ = ( $x - y)$ ( $x^{2} + xy + y^{2})$

$331$ = ( $x - y)$ ( $x^{2} + xy + y^{2})$

Thus, we can conclude that $x-y$ = 1 and ( $x^{2} + xy + y^{2})$ = 331. since 331 has only two factors (well it is a prime) , 1 and 331 $x-y = 1$

$x = y + 1$ .

Substitute $x=y + 1$ in to ( $x - y)$ ( $x^{2} + xy + y^{2})$ and you get $y^{2} + 2y + 1 + y^{2} + y + y^{2} = 331$ . (After expanding)

Regroup the factors to the left side

$3y^{2} + 3y - 330$ = 0.

$y^{2} + y - 110$ = 0

Solving the quadratic and you get $y = 10$ as the positive root.

Since $x = y + 1$ , $x = 11$ Yay you found the answer! Have a cookie🍪

$x \geq 7$ , $y = x - 1$

When $x = 8$ and $y = 7$ - $8^3 - 7^3 = 169$

...

When $x = 11$ and $y = 10$ - $11^3 - 10^3 = 331$

Therefore, $x = 11$