Total number of factors of 73^3 - 40^3 - 33^3 is?

Steps to Solve

1. Identify the expression We need to simplify the expression $73^3 - 40^3 - 33^3$.

2. Use the identity for differences of cubes Recall the identity: $$ a^3 - b^3 = (a-b)(a^2 + ab + b^2 $$. We can rewrite the expression as: $$ 73^3 - (40^3 + 33^3) $$.

3. Simplify $40^3 + 33^3$ using the sum of cubes Let $a = 40$ and $b = 33$. Then: $$ a^3 + b^3 = (a+b)(a^2 - ab + b^2) $$. Calculate $a + b = 40 + 33 = 73$ and $a^2 - ab + b^2 = 40^2 - (40)(33) + 33^2$.

4. Calculate $a^2 - ab + b^2$ Calculate each term:

• $40^2 = 1600$
• $33^2 = 1089$
• $ab = 40 \times 33 = 1320$

So, $$ a^2 - ab + b^2 = 1600 - 1320 + 1089 = 1369 $$.

5. Now substitute back into the expression Thus, $$ 73^3 - (40^3 + 33^3) = 73^3 - (73 \times 1369) = 73(73^2 - 1369) $$. Now calculate $73^2 - 1369 = 5329 - 1369 = 3960$, leading to: $$ 73 \times 3960 $$.

6. Find the prime factorization of $3960$ Break $3960$ down: $$ 3960 = 2^3 \times 3^2 \times 5 \times 11 $$.

7. Combine the factors So, $$ 73^1 \times 2^3 \times 3^2 \times 5^1 \times 11^1 $$.

8. Use the formula for total factors If $n = p_1^{e_1} \times p_2^{e_2} \times ...$, then the total number of factors is given by: $$(e_1 + 1)(e_2 + 1)...$$.

9. Calculate number of factors from the prime factorization Thus, for $73^1 \times 2^3 \times 3^2 \times 5^1 \times 11^1$:

• For $73^1$: $1 + 1 = 2$
• For $2^3$: $3 + 1 = 4$
• For $3^2$: $2 + 1 = 3$
• For $5^1$: $1 + 1 = 2$
• For $11^1$: $1 + 1 = 2$

Therefore: $$ 2 \times 4 \times 3 \times 2 \times 2 = 96 $$.