If frac{d}{dx} left( frac{x^4 + x^2 + 1}{x^2 | vedclass

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(A) Given the expression $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right)$.First,simplify the fraction inside the derivative:$\frac{x^4

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$3$

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(A) Given the expression $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right)$.First,simplify the fraction inside the derivative:$\frac{x^4 + x^2 + 1}{x^2 + x + 1} = \frac{(x^2 + x + 1)(x^2 - x + 1)}{x^2 + x + 1} = x^2 - x + 1$.Now,differentiate the simplified expression with respect to $x$:$\frac{d}{dx} (x^2 - x + 1) = 2x - 1$.Comparing $2x - 1$ with $ax + b$,we get $a = 2$ and $b = -1$.Therefore,$a - b = 2 - (-1) = 2 + 1 = 3$.

If $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right) = ax + b$,then $a - b =$

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