Question:

Step 1: Simplify the first fraction. The first fraction is $\dfrac{a+1}{a^{2/3} - a^{1/3} + 1}$. Recognize that the numerator $a+1$ can be factored as a sum of cubes. Let $x = a^{1/3}$, then $a+1 = (a^{1/3})^3 + 1^3$. Using the sum of cubes formula $x^3+y^3 = (x+y)(x^2-xy+y^2)$, we have: $$ a+1 = (a^{1/3} + 1)((a^{1/3})^2 - a^{1/3} \cdot 1 + 1^2) = (a^{1/3} + 1)(a^{2/3} - a^{1/3} + 1) $$ Therefore, the first fraction simplifies to: $$ \frac{(a^{1/3} + 1)(a^{2/3} - a^{1/3} + 1)}{a^{2/3} - a^{1/3} + 1} = a^{1/3} + 1 $$ Step 2: Simplify the second fraction. The second fraction is $\dfrac{a-1}{a - a^{1/2}}$. Factor the numerator using the difference of squares formula $x^2-y^2 = (x-y)(x+y)$. Let $x = a^{1/2}$, then $a-1 = (a^{1/2})^2 - 1^2$. $$ a-1 = (a^{1/2} - 1)(a^{1/2} + 1) $$ Factor the denominator by taking out the common term $a^{1/2}$: $$ a - a^{1/2} = a^{1/2}(a^{1/2} - 1) $$ Therefore, the second fraction simplifies to: $$ \frac{(a^{1/2} - 1)(a^{1/2} + 1)}{a^{1/2}(a^{1/2} - 1)} = \frac{a^{1/2} + 1}{a^{1/2}} = 1 + \frac{1}{a^{1/2}} = 1 + a^{-1/2} $$ Step 3: Combine the simplified fractions. The expression inside the parenthesis is the difference of the two simplified fractions: $$ (a^{1/3} + 1) - (1 + a^{-1/2}) = a^{1/3} + 1 - 1 - a^{-1/2} = a^{1/3} - a^{-1/2} $$ The problem requires finding the term independent of $a$ in the expansion of $\left(a^{1/3} - a^{-1/2}\right)^{10}$. Step 4: Determine the general term of the binomial expansion. The general term $T_{k+1}$ in the binomial expansion of $(x+y)^n$ is given by $T_{k+1} = \binom{n}{k} x^{n-k} y^k$. In this case, $x = a^{1/3}$, $y = -a^{-1/2}$, and $n = 10$. $$ T_{k+1} = \binom{10}{k} (a^{1/3})^{10-k} (-a^{-1/2})^k $$ $$ T_{k+1} = \binom{10}{k} a^{\frac{10-k}{3}} (-1)^k a^{-\frac{k}{2}} $$ Combine the powers of $a$: $$ T_{k+1} = \binom{10}{k} (-1)^k a^{\frac{10-k}{3} - \frac{k}{2}} $$ Step 5: Find the value of $k$ for the term independent of $a$. For the term to be independent of $a$, the exponent of $a$ must be zero. $$ \frac{10-k}{3} - \frac{k}{2} = 0 $$ Multiply the entire equation by 6 to eliminate the denominators: $$ 2(10-k) - 3k = 0 $$ $$ 20 - 2k - 3k = 0 $$ $$ 20 - 5k = 0 $$ $$ 5k = 20 $$ $$ k = 4 $$ Step 6: Calculate the term independent of $a$. Substitute $k=4$ into the general term formula: $$ T_{4+1} = T_5 = \binom{10}{4} (-1)^4 a^{\frac{10-4}{3} - \frac{4}{2}} $$ $$ T_5 = \binom{10}{4} (1) a^{\frac{6}{3} - 2} $$ $$ T_5 = \binom{10}{4} a^{2 - 2} $$ $$ T_5 = \binom{10}{4} a^0 $$ $$ T_5 = \binom{10}{4} $$ Calculate the binomial coefficient: $$ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} $$ $$ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24} = 10 \times 3 \times 7 = 210 $$ The term independent of $a$ is 210.