If frac{x^{3}-1}{p(x)}=frac{x^{2}+x+1}{x-1},t | vedclass

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(A) Given the equation: $\frac{x^{3}-1}{p(x)} = \frac{x^{2}+x+1}{x-1}$.We know the algebraic identity for the difference of cubes: $x^{3}-1 = (x-1)(x^

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$(x-1)^{2}$

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(A) Given the equation: $\frac{x^{3}-1}{p(x)} = \frac{x^{2}+x+1}{x-1}$.We know the algebraic identity for the difference of cubes: $x^{3}-1 = (x-1)(x^{2}+x+1)$.Substituting this into the equation: $\frac{(x-1)(x^{2}+x+1)}{p(x)} = \frac{x^{2}+x+1}{x-1}$.By canceling $(x^{2}+x+1)$ from both sides (assuming $x^{2}+x+1 
eq 0$): $\frac{x-1}{p(x)} = \frac{1}{x-1}$.Cross-multiplying gives: $p(x) = (x-1)(x-1) = (x-1)^{2}$.

If $\frac{x^{3}-1}{p(x)}=\frac{x^{2}+x+1}{x-1}$,then $p(x) = \dots$

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