Note that the expression \( rac{t^3 - 8}{t - 2} \) is indeterminate at \( t = 2 \) since both numerator and denominator vanish. Factor the numerator:

["# How to Simplify ( \frac{t^3 - 8}{t - 2} ): Understanding the Indeterminate Form at ( t = 2 )", "When working with rational expressions, one common challenge is identifying indeterminate forms—especially at points where both the numerator and denominator evaluate to zero. A classic example is the expression:", "[\n\frac{t^3 - 8}{t - 2}\n]", "At first glance, substituting ( t = 2 ) yields:", "Numerator: ( 2^3 - 8 = 8 - 8 = 0 )\nDenominator: ( 2 - 2 = 0 )", "This results in the indeterminate form ( \frac{0}{0} ), meaning we cannot determine the value directly from the expression as written. However, this form is not inherently undefined—it signals that the expression may simplify neatly. The key to resolving this is factoring the numerator, which reveals the underlying structure and removes the indeterminacy.", "## Factoring the Numerator", "The numerator ( t^3 - 8 ) is a difference of cubes, since ( 8 = 2^3 ). The difference of cubes formula states:", "[\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\n]", "Applying this with ( a = t ) and ( b = 2 ), we get:", "[\nt^3 - 8 = (t - 2)(t^2 + 2t + 4)\n]", "Thus, the original expression becomes:", "[\n\frac{t^3 - 8}{t - 2} = \frac{(t - 2)(t^2 + 2t + 4)}{t - 2}\n]", "For all values of ( t <br/>\ne 2 ), the ( t - 2 ) terms in the numerator and denominator cancel out, leaving:", "[\nt^2 + 2t + 4\n]", "## Resolving the Indeterminate Form", "Since ( t - 2 <br/>\ne 0 ) when ( t <br/>\ne 2 ), the expression simplifies to the polynomial ( t^2 + 2t + 4 ), which is defined everywhere. However, at ( t = 2 ), the original expression is technically undefined due to division by zero—hence the indeterminate form.", "But because the expression simplifies continuously (except at ( t = 2 )), we can define its value at ( t = 2 ) via the simplified expression:", "[\n\lim_{t \ o 2} \frac{t^3 - 8}{t - 2} = \lim_{t \ o 2} (t^2 + 2t + 4) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12\n]", "This limit represents the actual value the expression approaches as ( t ) approaches 2, resolving the indeterminacy.", "## Practical Implications and Summary", "Understanding indeterminate forms is crucial in calculus, algebra, and mathematical modeling. Factoring provides a clear pathway to simplify expressions and evaluate limits efficiently. For the expression:", "[\n\frac{t^3 - 8}{t - 2}\n]", "After factoring and canceling the common ( t - 2 ) factor (with ( t <br/>\ne 2 )), we obtain a smooth, continuous function that equals 12 at ( t = 2 ) by continuity.", "Key Takeaways:\n- ( \frac{t^3 - 8}{t - 2} ) is indeterminate at ( t = 2 ) because both numerator and denominator are zero.\n- Factoring the numerator using the difference of cubes reveals a removable discontinuity.\n- The simplified expression ( t^2 + 2t + 4 ) is defined everywhere and equals 12 at ( t = 2 ) by continuity.", "By factoring early, we resolve indeterminate forms with confidence and clarity—making calculations both mathematically rigorous and conceptually accessible.", "---", "Keywords:\nindeterminate form ( \frac{t^3 - 8}{t - 2} ), factor numerator, simplify rational expressions, limit at ( t = 2 ), difference of cubes, calculus, algebra, removable discontinuity."]