1. Key Concept: Binomial Theorem

The binomial theorem provides a formula for the algebraic expansion of powers of a binomial. For an expression of the form $(X+Y)^n$, the general term (or $(r+1)^{th}$ term) in its expansion is given by: $T_{r+1} = {}^{n}C_r X^{n-r} Y^r$ where ${}^nC_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient. This formula is crucial for finding specific terms or coefficients without expanding the entire expression.

2. Analyzing the First Expansion: $\left(ax^3 + \frac{1}{bx^{1/3}}\right)^{15}$

We want to find the coefficient of $x^{15}$ in the expansion of $\left(ax^3 + \frac{1}{bx^{1/3}}\right)^{15}$. Here, $n=15$, $X = ax^3$, and $Y = \frac{1}{bx^{1/3}} = b^{-1}x^{-1/3}$.

The general term, $T_{r+1}$, is: $T_{r+1} = {}^{15}C_r (ax^3)^{15-r} \left(\frac{1}{bx^{1/3}}\right)^r$

Now, let's separate the coefficient part and the $x$ part. $T_{r+1} = {}^{15}C_r a^{15-r} (x^3)^{15-r} (b^{-1})^r (x^{-1/3})^r$ $T_{r+1} = {}^{15}C_r a^{15-r} b^{-r} x^{3(15-r)} x^{-r/3}$ $T_{r+1} = {}^{15}C_r a^{15-r} b^{-r} x^{(45 - 3r - r/3)}$ $T_{r+1} = {}^{15}C_r a^{15-r} b^{-r} x^{(45 - \frac{10r}{3})}$

To find the coefficient of $x^{15}$, we equate the power of $x$ to $15$: $45 - \frac{10r}{3} = 15$ Subtract 15 from both sides: $30 = \frac{10r}{3}$ Multiply by 3 and divide by 10: $r = \frac{30 \times 3}{10}$ $r = 9$

Now, substitute $r=9$ back into the coefficient part of the general term: Coefficient of $x^{15} = {}^{15}C_9 a^{15-9} b^{-9} = {}^{15}C_9 a^6 b^{-9} = {}^{15}C_9 \frac{a^6}{b^9}$

3. Analyzing the Second Expansion: $\left(ax^{1/3} - \frac{1}{bx^3}\right)^{15}$

Next, we find the coefficient of $x^{-15}$ in the expansion of $\left(ax^{1/3} - \frac{1}{bx^3}\right)^{15}$. Here, $n=15$, $X = ax^{1/3}$, and $Y = -\frac{1}{bx^3} = -b^{-1}x^{-3}$.

The general term, $T_{k+1}$ (using $k$ to distinguish from the previous $r$), is: $T_{k+1} = {}^{15}C_k (ax^{1/3})^{15-k} \left(-\frac{1}{bx^3}\right)^k$

Separate the coefficient part and the $x$ part: $T_{k+1} = {}^{15}C_k a^{15-k} (x^{1/3})^{15-k} (-1)^k (b^{-1})^k (x^{-3})^k$ $T_{k+1} = {}^{15}C_k a^{15-k} (-1)^k b^{-k} x^{\frac{15-k}{3}} x^{-3k}$ $T_{k+1} = {}^{15}C_k a^{15-k} (-1)^k b^{-k} x^{\left(\frac{15-k}{3} - 3k\right)}$ $T_{k+1} = {}^{15}C_k a^{15-k} (-1)^k b^{-k} x^{\left(\frac{15-k-9k}{3}\right)}$ $T_{k+1} = {}^{15}C_k a^{15-k} (-1)^k b^{-k} x^{\left(\frac{15-10k}{3}\right)}$

To find the coefficient of $x^{-15}$, we equate the power of $x$ to $-15$: $\frac{15-10k}{3} = -15$ Multiply by 3: $15 - 10k = -45$ Add $10k$ to both sides and add 45 to both sides: $15 + 45 = 10k$ $60 = 10k$ $k = 6$

Now, substitute $k=6$ back into the coefficient part of the general term: Coefficient of $x^{-15} = {}^{15}C_6 a^{15-6} (-1)^6 b^{-6}$ Since $(-1)^6 = 1$: Coefficient of $x^{-15} = {}^{15}C_6 a^9 b^{-6} = {}^{15}C_6 \frac{a^9}{b^6}$

4. Equating the Coefficients

The problem states that the coefficient of $x^{15}$ in the first expansion is equal to the coefficient of $x^{-15}$ in the second expansion. Therefore: ${}^{15}C_9 \frac{a^6}{b^9} = {}^{15}C_6 \frac{a^9}{b^6}$

We know that ${}^nC_r = {}^nC_{n-r}$. So, ${}^{15}C_9 = {}^{15}C_{15-9} = {}^{15}C_6$. Since the binomial coefficients are equal, we can cancel them from both sides of the equation: $\frac{a^6}{b^9} = \frac{a^9}{b^6}$

Now, we solve for the relationship between $a$ and $b$. Since $a$ and $b$ are positive real numbers, we know $a, b \neq 0$. Divide both sides by $a^6$: $\frac{1}{b^9} = \frac{a^3}{b^6}$ Multiply both sides by $b^9$: $1 = a^3 \frac{b^9}{b^6}$ $1 = a^3 b^3$ This can be written as: $(ab)^3 = 1$ Since $a$ and $b$ are positive real numbers, $ab$ must also be a positive real number. The only positive real number whose cube is 1 is 1 itself. Therefore: $ab = 1$

5. Tips and Common Mistakes

• General Term Formula: Always start by correctly writing down the general term $T_{r+1} = {}^{n}C_r X^{n-r} Y^r$. Pay close attention to the powers of $X$ and $Y$.
• Sign Convention: If the binomial is $(X-Y)^n$, remember that the second term is $-Y$. So, $Y^r$ becomes $(-Y)^r = (-1)^r Y^r$. This is crucial for getting the correct sign of the coefficient.
• Fractional and Negative Exponents: Be careful while manipulating exponents, especially when dealing with fractions or negative signs. Combine them correctly to get the net power of $x$.
• Binomial Coefficient Property: Remember the property ${}^nC_r = {}^nC_{n-r}$. This often helps simplify equations when equating coefficients.
• Algebraic Simplification: After equating coefficients, carefully simplify the algebraic expression to find the relationship between the variables. Don't rush through this step.

6. Summary and Key Takeaway

By correctly applying the binomial theorem and determining the general terms for both expansions, we found the values of $r$ that yield the desired powers of $x$. Equating the coefficients derived from these terms led to the relationship $(ab)^3 = 1$. Given that $a$ and $b$ are positive real numbers, the final solution is $ab=1$. This problem emphasizes a thorough understanding of the binomial theorem's general term and careful algebraic manipulation of exponents and coefficients.