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

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(B) Given equation: $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$.Since $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$,we can factor out $(x^2 +

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

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(B) Given equation: $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$.Since $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$,we can factor 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 $x^2 + x + 1 = 0$ has discriminant $D = 1 - 4 = -3 Thus,the real roots must come from $ax^2 + x + a = 0$.For the roots to be real and distinct,we require $a 
eq 0$ and 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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