If (x+3) and (x^2+3) are perfect cubes, then so is (x+3)(x^2+3). However,
(x+3)(x^2+3) = x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2. Since 3^3-2^3 = 19 > 2, no two cubes can be only 2 apart, so the above has no solution.

Ax

Your algebra is a bit off, (x+3)(x^2+3) = (x+1)^3+8.
The only cubes that differ by 8 are 0 and +/-8, which leaves you with non-integer values for a and/or b, thus no solution exists.

Use the same argument, but conclude with “If the product of the two cubes (still a cube) is equal to (x+1)^3 + 8 = (x+1)^3 + 2^3, then there would exist a triple satisfying Fermat’s last Theorem, which as shown by Wiles, is impossible.

Charon l'CypherIf (x+3) and (x^2+3) are perfect cubes, then so is (x+3)(x^2+3). However,

(x+3)(x^2+3) = x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2. Since 3^3-2^3 = 19 > 2, no two cubes can be only 2 apart, so the above has no solution.

AxYour algebra is a bit off, (x+3)(x^2+3) = (x+1)^3+8.

The only cubes that differ by 8 are 0 and +/-8, which leaves you with non-integer values for a and/or b, thus no solution exists.

BGroninUse the same argument, but conclude with “If the product of the two cubes (still a cube) is equal to (x+1)^3 + 8 = (x+1)^3 + 2^3, then there would exist a triple satisfying Fermat’s last Theorem, which as shown by Wiles, is impossible.

CommanderHanaYou, sir, just made my day.