Find two numbers such that their difference and also the difference of their cubes are given numbers; say, their difference is 6 and the difference of their cubes are 504. Call the numbers $x + 3$ and $x - 3$.
2026-04-06 20:05:56.1775505956
Diophantus math
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Well, using the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2),$ we get $$\begin{align}504 &= (x+3)^3-(x-3)^3\\ &= \bigl((x+3)-(x-3)\bigr)\left((x+3)^2+(x+3)(x-3)+(x-3)^2\right)\\ &= 6\bigl((x^2+6x+9)+(x^2-9)+(x^2-6x+9)\bigr)\\ &= 6(3x^2+9)\\ &= 18x^2+54.\end{align}$$ You can take it from there, yes?