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
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
341
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)}