Demonstrate that (1-x1)(1-x2)(1-x3)(1-x4) = 5 where x1,x2,x3,x4 roots of the equation x^4 + x^3 + x^2 + x + 1 = 5
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We have the equation:
x^4+x^3+x^2+x+1=5
x^4+x^3+x^2+x-4=0
According Viete's rule we know that:
x1+x2+x3+x4=-b/a= -1
x1*x2+x1*x3+x1*x4+x2*x3+x2*x4+x3*x4=1
x1*x2*x3+x1*x2*x4+x1*x3*x4+ x2*x3*x4=1
x1*x2*x3*x4=1
We need to prove that:
(1-x1)(1-x2)(1-x3)(1-x4)=5
Let us open the brackets:
(1-x1)(1-x2)(1-x3)(1-x4)=1-(x1 + x2 + x3 + x4) + (x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4)-(x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4)+(x1*x2*x3*x4)
= 1+1+1+1+1=5
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x^4+x^3+x^2+x+1 = 5.
Then to show that (1-x1)(1-x2)(1-x3)(1-x4) = 5, if x1,x2 ,x3 and x4 are the roots of x^4+x^3+x^2+x+1 = 5
Solution:
If x1, x2,x3 and x4 are the roots of x^4+x^3+x^2+x+1 = 5 Or x^4+x^3+x^2+x-4 =0, then we can write:
x^4+x^3+x^2+x-4 = (x-x1)(x-x2)(x-x3)(x-x4). Now since this should be identity. so putting x = 1, we get:
1+1+1+1-4 = (x1-x1)(1-x2)(1-x3)(1-x4) Or
(1-x1)(1-x2)(1-x3)(1-x4) = 0 and not 5.
We'll write Viete expressions:
x1 + x2 + x3 + x4 =-1
x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4=1
x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4=1
x1*x2*x3*x4=1
We'll open the brackets and we'll get:
(1-x1)(1-x2)(1-x3)(1-x4)=1-(x1 + x2 + x3 + x4) + (x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4) - (x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4) + (x1*x2*x3*x4)
We'll substitute by the results from the Viete's relations:
(1-x1)(1-x2)(1-x3)(1-x4)= 1+1+1+1+1=5
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