If the coefficient of x^15 in the expansion of (ax^3+1bx^1/3)^15 is equal to the coefficie…

Question

If the coefficient of $x^{15}$ in the expansion of $\left(ax^3+\dfrac{1}{bx^{1/3}}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $\left(ax^{1/3}-\dfrac{1}{bx^3}\right)^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(a,b)$:

Accepted Answer

For $\left(ax^3+\dfrac{1}{bx^{1/3}}\right)^{15}$: $T_{r+1}=\binom{15}{r}a^{15-r}b^{-r}x^{3(15-r)-r/3}$. Power: $45-3r-r/3=45-\frac{10r}{3}$. Set $=15$: $r=9$. Coefficient $={}^{15}C_9a^6b^{-9}$.

For $\left(ax^{1/3}-\dfrac{1}{bx^3}\right)^{15}$: $T_{r+1}=\binom{15}{r}a^{15-r}(-b)^{-r}x^{(15-r)/3-3r}$. Power: $5-r/3-3r=5-\frac{10r}{3}$. Set $=-15$: $r=6$. Coefficient $={}^{15}C_6a^9b^{-6}$.

Equating (${}^{15}C_9={}^{15}C_6$): $a^6b^{-9}=a^9b^{-6}\Rightarrow a^3b^3=1\Rightarrow ab=1$

Solutions · 1