2

Let $a,b,c,d,e$ be five numbers satisfying the following conditions: $$a+b+c+d+e =0$$ and $$abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=33$$ Find the value of $$\frac{a^3+b^3+c^3+d^3+e^3}{502}$$

My Approach:

$$(a+b+c+d+e)^3 = \sum_{a,b,c,d,e}{a^3} + 3\sum_{a,b,c,d,e}{a^2b} + 6\sum_{a,b,c,d,e}{abc} $$

Taking $\mod (a+b+c+d+e)$, $$(a+b) ≡ -(c+d+e)$$ $$ab(a+b) ≡ -ab(c+d+e)$$ $$\sum{a^2b} ≡ -ab(c+d+e) -bc(a+d+e) -cd(a+b+e)-... = -\sum_{a,b,c,d,e}{ab(c+d+e)} = -3\sum_{a,b,c,d,e}{abc}$$ Therefore, $\sum{a^2b} = p(a,b,c,d,e) . (a+b+c+d+e) - 3\sum_{a,b,c,d,e}{abc}$

Since, $(a+b+c+d+e) = 0$ $$\sum{a^3} = (3×3 -6)\sum{abc} = 3×33 = \color{red}{99}$$

But the answer key shows: $$\frac{\sum{a^3}}{\color{blue}{502}} = 99$$

Where is my mistake?

ACB
  • 3,713
user961447
  • 319

1 Answers1

4

OP's result is correct. For verification, Newton's identity $\,p_3=e_1^3-3e_1e_2+3e_3\,$ for the sum of cubes gives the same result directly (where the sums are the symmetric sums over $a,b,c,d,e$):

$$ \sum a^3 = \left(\sum a\right)^3 - 3\,\left(\sum a\right)\left(\sum ab\right) + 3 \left(\sum abc\right) = 0 - 3 \cdot 0 \cdot \left(\sum ab\right) + 3 \cdot 33 = 99 $$

dxiv
  • 76,497