If the two roots of the equation (a - 1)(x^4 | vedclass

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(B) Given equation: $(a-1)(x^4+x^2+1) + (a+1)(x^2+x+1)^2 = 0$We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Substituting this,we get: $(a-1)(x^2+x+1)(x

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$(-1/2, 0) \cup (0, 1/2)$

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(B) Given equation: $(a-1)(x^4+x^2+1) + (a+1)(x^2+x+1)^2 = 0$We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Substituting this,we get: $(a-1)(x^2+x+1)(x^2-x+1) + (a+1)(x^2+x+1)^2 = 0$Factoring out $(x^2+x+1)$: $(x^2+x+1)[(a-1)(x^2-x+1) + (a+1)(x^2+x+1)] = 0$Simplifying the expression inside the bracket: $(x^2+x+1)[ax^2-ax+a-x^2+x-1 + ax^2+ax+a+x^2+x+1] = 0$$(x^2+x+1)(2ax^2+2x+2a) = 0$$2(x^2+x+1)(ax^2+x+a) = 0$The quadratic $x^2+x+1$ has a discriminant $D = 1^2 - 4(1)(1) = -3 For the equation to have real and distinct roots,the quadratic $ax^2+x+a = 0$ must have two distinct real roots.This requires $a 
eq 0$ and the discriminant $D > 0$.$D = 1^2 - 4(a)(a) = 1 - 4a^2 > 0$$4a^2 Since $a 
eq 0$,the set of values is $a \in (-1/2, 0) \cup (0, 1/2)$.

If the two roots of the equation $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$ are real and distinct,then the set of all values of $a$ is

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