4a = -b
=> b2 = 3
=> b =± √3 & a = ±√3
Find a and b given that
(x^6+1) = (x^2+1)(x^2+ax+1)(x^2+bx+1)
is an identity
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11 Answers
4RHS = x6 + ( a + b)x5 + (3 + ab)x4 + (2a+2b)x3 + (ab+3)x2 + (a+b)x + 1
Equating coefficients of LHS & RHS
a + b = 0
ab = -3
341maybe she followed the link from the homepage and so didnt notice.
But there's an easier method than comparing coefficients. So fresh attempts are welcome
11prophet sir,
is it solving the equations
(2+a)(2+b)=1 & (2-a)(2-b)=1......
24yeah soumik did by best way...
since identity ,so infinite soln..
putting x=1
1=(a+2)(b+2)
putting x=-1
1=(2-a)(2-b)
1yes ..
but hey why are we trying to steal the thunder from some juniors :D
1Instead of going for different methods can't we just use the basic
x6+1= (x2)3+13
0n solving we get
RHS= (x2+1)(x2+√3x+1)(x2-√3x+1)
Hence we get the solution set
(a,b) ={(√3,-√3);(-√3,√3)}