Rationalizing the Numerator to Evaluate a Limit for a Tough Function

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To evaluate the limit as x approaches a for the expression (x^(1/3) - a^(1/3))/(x - a), it's essential to factor out (x - a) from the numerator. This can be achieved using the identity for the difference of cubes, where A = x^(1/3) and B = a^(1/3), leading to the factorization (x - a) = (x^(1/3) - a^(1/3))(x^(2/3) + a^(1/3)x^(1/3) + a^(2/3)). Alternatively, one can rationalize the numerator by multiplying both the numerator and denominator by the conjugate, which also simplifies the limit evaluation. This approach allows the limit to be computed without the denominator equating to zero.
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Evaluate lim as x approaches a for (x^1/3 - a^1/3)/(x-a). I want to factor out x-a from the numerator so that the denominator is not equal to zero. How can I do this?
 
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(A^3-B^3)=(A-B)(A^2+AB+B^2). Use that formula where A=x^(1/3) and B=a^(1/3). I.e. factor (x-a).
 
Another way to do that (Dick's hint is perfectly good) is not to factor but to multiply to rationalize the numerator. Using the same formula Dick gave, x- a= (x1/3- a1/3)(x2/3+ a1/3x1/3+ a2/3). Multiply both numerator and denominator by that.
 
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