Past, present, and future again
Given
E ( a , b , c ) = a ( b 3 − c 3 ) + b ( c 3 − a 3 ) + c ( a 3 − b 3 ) ,
find the value of
( 2 0 1 4 − 2 0 1 6 ) ( 2 0 1 4 + 2 0 1 6 + 2 0 1 8 ) E ( 2 0 1 4 , 2 0 1 6 , 2 0 1 8 ) .
The answer is -8.
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2 solutions
Interesting, I didn't think of it that way. What I saw is that E ( b , b , c ) = 0 → ( a − b ) is a factor, and since E ( a , b , c ) is cyclic, then ( a − b ) ( b − c ) ( c − a ) must also be a factor. Since the degree of E ( a , b , c ) is 4 but the degree of ( a − b ) ( b − c ) ( c − a ) is 3, then E ( a , b , c ) = ( a − b ) ( b − c ) ( c − a ) k ( a + b + c ) . By substituting some values for a , b , c , we can find that k = 1 and thus the fully factorized form of E ( a , b , c ) = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) . We cancel ( a − b ) ( a + b + c ) in the denominator, and this leaves ( b − c ) ( c − a ) , and since b = 2 0 1 6 , c = 2 0 1 8 , a = 2 0 1 4 , this gives − 8
Wow...never thought determinants could be used to solve this
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Yep.... Discriminants make life easy... :-)
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Only if you know how to use it. If you're not good with it, things become a mess very quickly.
For these types of problems, since the values ( 2 0 1 4 , 2 0 1 6 , 2 0 1 8 ) are given, I would just prefer to substitute ( n − 2 , n , n + 2 ) and simplify it
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@Hung Woei Neoh – That's what also striked me first time but as usual I was lazy to grab pen and paper so I converted into determinants and then applied properties to simplify....
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@Rishabh Jain – Lol. Ever since I started my Brilliant journey, I always keep a notebook and a pencil with me wherever I am, so that I can start solving problems whenever I feel like it. Of course, I only type solutions with my laptop
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@Hung Woei Neoh – That's a very good habit... :-)
you mean determinants
Let a = 2 0 1 4 = n − 2 , b = 2 0 1 6 = n , c = 2 0 1 8 = n + 2
( 2 0 1 4 − 2 0 1 6 ) ( 2 0 1 4 + 2 0 1 6 + 2 0 1 8 ) E ( 2 0 1 4 , 2 0 1 6 , 2 0 1 8 ) = ( n − 2 − n ) ( n − 2 + n + n + 2 ) ( n − 2 ) ( n 3 − ( n + 2 ) 3 ) + n ( ( n + 2 ) 3 − ( n − 2 ) 3 ) + ( n + 2 ) ( ( n − 2 ) 3 − n 3 ) = − 2 ( 3 n ) ( n − 2 ) ( n 3 − n 3 − 6 n 2 − 1 2 n − 8 ) + n ( n 3 + 6 n 2 + 1 2 n + 8 − n 3 + 6 n 2 − 1 2 n + 8 ) + ( n + 2 ) ( n 3 − 6 n 2 + 1 2 n − 8 − n 3 ) = − 6 n ( n − 2 ) ( − 6 n 2 − 1 2 n − 8 ) + n ( 1 2 n 2 + 1 6 ) + ( n + 2 ) ( − 6 n 2 + 1 2 n − 8 ) = − 6 n − 6 n 3 + 1 2 n 2 − 1 2 n 2 + 2 4 n − 8 n + 1 6 + 1 2 n 3 + 1 6 n − 6 n 3 − 1 2 n 2 + 1 2 n 2 + 2 4 n − 8 n − 1 6 = − 6 n 4 8 n = − 8
Nice... (+1)
See Determinants E ( a , b , c ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ a a 3 1 b b 3 1 c c 3 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ Perform C 1 → C 1 − C 2 ; C 2 → C 2 − C 3
= ( a − b ) ( b − c ) ∣ ∣ ∣ ∣ ∣ ∣ 1 a 2 + a b + b 2 0 1 b 2 + b c + c 2 0 c c 3 1 ∣ ∣ ∣ ∣ ∣ ∣
Now simply expand along R 3
E ( a , b , c ) = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ⟹ ( b − c ) ( a + b + c ) E ( a , b , c ) = ( a − b ) ( c − a )
Put a = 2 0 1 4 , b = 2 0 1 6 , c = 2 0 1 8
⟹ ( 2 0 1 4 − 2 0 1 6 ) ( 2 0 1 4 + 2 0 1 6 + 2 0 1 8 ) E ( 2 0 1 4 , 2 0 1 6 , 2 0 1 8 ) = − 8