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Dear Ajinkya
x6 - 10x3 + 27=0
let x3 =t
so t2 -10t +27 =0
discriminant =(-10)^2-4(1)(27) =-8 which is less than 0.
So,discriminant of this equation is negative so no real value of t.
So complex roots exist for this equation,which are {10+i(8)^(1/2)}/2 and {10-i(8)^(1/2)}/2.
So this equation has no real value of x.
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