Cube Roots of Unity
If ( a + ω ) − 1 + ( b + ω ) − 1 + ( c + ω ) − 1 + ( d + ω ) − 1 = 2 ω − 1 and ( a + ω ′ ) − 1 + ( b + ω ′ ) − 1 + ( c + ω ′ ) − 1 + ( d + ω ′ ) − 1 = 2 ω ′ − 1 ; where ω and ω ′ are the complex cube roots of unity, then what is the value of ( a + 1 ) − 1 + ( b + 1 ) − 1 + ( c + 1 ) − 1 + ( d + 1 ) − 1 ?
The answer is 2.
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Consider the equation
( a + x ) − 1 + ( b + x ) − 1 + ( c + x ) − 1 + ( d + x ) − 1 = 2 x − 1 .
Upon simplifying and writing in the standard form, it becomes
2 x 4 + ( a + b + c + d ) x 3 − ( a b c + a b d + a c d + b c d ) x − 2 a b c d = 0 .
It is given that ω and ω ′ are two of the roots. Let the other two roots be R 1 and R 2 . Then the sum of the products of the roots taken two at a time is equal to zero (since the coefficient of x 2 in the last equation is zero), i.e.
ω ω ′ + R 1 ω + R 1 ω ′ + R 2 ω + R 2 ω ′ + R 1 R 2 = 0
or, 1 − R 1 − R 2 + R 1 R 2 = 0
or, ( 1 − R 1 ) ( 1 − R 2 ) = 0
Therefore, R 1 = R 2 = 1 .
It follows that whenever ω and ω ′ are the roots of the original equation, then 1 is also a root. Hence putting x = 1 in the equation, we get
( a + 1 ) − 1 + ( b + 1 ) − 1 + ( c + 1 ) − 1 + ( d + 1 ) − 1 = 2