Verified answer

For any integer n we have n³ ≡ 0, 1, or 6 (mod 7).

Hence

a³ + b³ ≡ 0, 1, 2, 5, or 6 (mod 7),

while

7c³ + 3 ≡ 3 (mod 7).

Thus there are no a,b,c ∈ ℤ satisfying the given equation.

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What specifically don't you understand?

Sometimes I wish we could downvote question-askers. Suleiman did explain clearly, which is ironically more than I can say for the OP's request for clarification. He used modular arithmetic--essentially, he wondered what remainder you get if you divide a^3 + b^3 and 7c^3+3 by 7. 7c^3+3 gives a remainder of 3, but his computation shows that a^3+b^3 gives only remainders of 0, 1, 2, 5 or 6, so there can't be any solution.

Two invalid statements do not necessarily add to make a third.

Example: Johns age is 4 and Mary's age is 6 so their total age is 10

John's age is not 2

Mary's age is not 8

Adding these negatives to get total age is not 10 would be wrong

But perhaps adding statements not valid for ANY integer would be OK

a^3 + b^3 = c^3 not valid for any integer

0 = 6c^3 + 3 not valid for any integer. (One real non-integer and 2 complex conjugate roots).

Plan A was to add these, but I leave it to you to decide whether these two wrongs really make a right ?

Regards - Ian