Elegant Eliminant

Algebra Level 4

Eliminate x x from:

x 2 + x = a x^{2}+x=a

x 3 = b x^{3}=b

If the eliminant is of the form:

C 0 a 3 + C 1 a 2 b + C 2 a b 2 + C 3 b 3 C_{0}a^{3}+C_{1}a^{2}b+C_{2}ab^{2}+C_{3}b^{3} + C 4 a 2 + C 5 a b + C 6 b 2 + C 7 a + C 8 b + C 9 = 0 +C_{4}a^{2}+C_{5}ab+C_{6}b^{2}+C_{7}a+C_{8}b+C_{9}=0

Compute: ∣ ∑ i = 0 9 C i 2 9 − i ∣ |\sum_{i=0}^9 C_{i}2^{9-i}|

NOTE: I) All C i C_{i} s are coprime. II) C i C_{i} s aren't binomial coefficients; Just arbitrary coefficients


The answer is 454.

Harkirat Dhanoa by Harkirat Dhanoa , 22, India

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1 solution

Harkirat Dhanoa
Feb 21, 2016

Recall the formula
( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)

Now, cubing both sides of x 2 + x = a x^{2}+x=a

We've x 6 + x 3 + 3 x 2 x ( x 2 + x ) = a 3 x^{6}+x^{3}+3x^{2}x(x^{2}+x)=a^{3}

i.e. b 2 + b + 3 b a = a 3 b^{2}+b+3ba=a^{3} (follows from the given equations)

Therefore C 0 = 1 C_{0}=1 , C 5 = − 3 C_{5}=-3 , C 6 = − 1 C_{6}=-1 , and C 8 = − 1 C_{8}=-1 rest of the coefficients being zero.

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