Prove that 369^3 - 219^3 is divisible by 1350.

Step-by-step solution

1. 1

   Factor with the difference-of-cubes identity instead of computing the powers. Using $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$ with $a=369$, $b=219$:

   $369^3-219^3=(369-219)\left(369^2+369\cdot 219+219^2\right)=150\left(369^2+369\cdot 219+219^2\right)$

   The first factor already supplies a large chunk of $1350$.

2. 2

   Compare $150$ with the target $1350$. Since

   $1350=150\cdot 9$

   the whole proof reduces to showing that the second factor is divisible by $9$. Framing the goal this way is what turns a large computation into a small one.

3. 3

   Show the second factor is a multiple of 9. Both $369$ and $219$ are divisible by $3$: $369=3\cdot 123$ and $219=3\cdot 73$. Therefore every term in $369^2+369\cdot 219+219^2$ carries a factor $3^2=9$:

   $369^2+369\cdot 219+219^2=9\left(123^2+123\cdot 73+73^2\right)$

   so the second factor is $9$ times an integer, with no need to evaluate anything.

4. 4

   Assemble the conclusion.

   $369^3-219^3=150\cdot 9\left(123^2+123\cdot 73+73^2\right)=1350\left(123^2+123\cdot 73+73^2\right)$

   which is $1350$ times an integer, proving the divisibility.

5. 5

   Confirm with the actual numbers. The inner sum is $15129+8979+5329=29437$, so

   $369^3-219^3=1350\cdot 29437=39\,739\,950$

   Direct computation agrees: $369^3=50\,243\,409$ and $219^3=10\,503\,459$, and their difference is $39\,739\,950\;\checkmark$, with $39\,739\,950\div 1350=29\,437$ exactly.