Question 16

Algebra Level 5

( a + b ω + c ω 2 ) 3 + ( a + c ω + b ω 2 ) 3 = 0 \large ( a + b\omega + c\omega^{2})^{3} +(a + c\omega + b\omega^{2})^{3}=0

If a , b a,b and c c are distinct integers in a geometric progression which satisfy the equation above where ω \omega is a non-real cube root of unity, what is the least possible value of ∣ a ∣ + ∣ b ∣ + ∣ c ∣ |a| + |b| + |c| ?


The answer is 7.

Samarpit Swain by Samarpit Swain , 23, India

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1 solution

Parv Jain
Apr 19, 2016

Let a, b, c respectively be a, ar, ar^2. The given expression simplifies to a^3((1+ rw+ r^2w^2)^3+ (1+ rw^2+ r^2w)^3)=0. Applying the formula of x^3+ y^3. We get; 2-r- r^2=0. r=1 or -2. r can not be 1 as a, b, c are distinct. Hence sum reduces to a(1+2+7) =7a. Minmum value 7.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Did you really use x^3+y^3?I used sum of G.P in both parenthesis and the numerator was same so cancelled. then added the 2 fractions and solved to get r=-2.

Ajinkya Shivashankar - 4 years, 7 months ago

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