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If x2 + (1/x2) = 1 then what is the value of x48 + x42 + x36 + x30 + x24 + x18 + x12 + x6 + 1 ?
- A-9
- B0
- C1
- D9
Answer not set
Discussion 1
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0
CCommunity
Given: x2 + (1/x2) = 1
Adding 2 on both sides
x2 + (1/x2) + 2 = 1 + 2
x2 + (1/x2) + 2(x)(1/x) = 1 + 2
(x + 1/x)2 = 3
(x + 1/x) = √3
Cubing both sides
(x + 1/x)3 = (√3)3
x3 + 1/x3 + 3(x)(1/x)(x + 1/x) = 3√3
x3 + 1/x3 = 0
x3 = -1/x3
x6 = -1
Putting values of x6 = -1 in x48 + x42 + x36 + x30 + x24 + x18 + x12 + x6 + 1
Answer = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 = 1
Adding 2 on both sides
x2 + (1/x2) + 2 = 1 + 2
x2 + (1/x2) + 2(x)(1/x) = 1 + 2
(x + 1/x)2 = 3
(x + 1/x) = √3
Cubing both sides
(x + 1/x)3 = (√3)3
x3 + 1/x3 + 3(x)(1/x)(x + 1/x) = 3√3
x3 + 1/x3 = 0
x3 = -1/x3
x6 = -1
Putting values of x6 = -1 in x48 + x42 + x36 + x30 + x24 + x18 + x12 + x6 + 1
Answer = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 = 1