If y - x - 1x then x4 - x3 - 4x2 - x - 1 - 0 becomes

The correct option is E x^2(y^2 + y 6) = 0 y=x+ 1 x y ^ 2 = x ^ 2 + 1 x ^ 2 +2 ⟹ x ^ 4 + x ^ 3 4 x ^ 2 +x+1=0 ⟹ x ^ 2 ...

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Multiplication of Any Polynomial

If y = x + ...

Question

If $y = x + \frac{1}{x}$ then $x^{4} + x^{3} - 4 x^{2} + x + 1 = 0$ becomes

x^{2} (y^{2} + y - 2) = 0

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x^{2} (y^{2} + y - 3) = 0

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x^{2} (y^{2} + y - 4) = 0

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x^{2} (y^{2} + y - 6) = 0

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none of these

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x = a^{1 / 2} + a^{- 1 / 2}, y = a^{1 / 2} - a^{- 1 / 2}

(x^{4} - x^{2} y^{2} - 1) + (y^{4} - x^{2} y^{2} + 1)

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