Binomial Theorem Question - None

Question

If the coefficient of x¹⁵ in the expansion of $\left(a x^3+\frac{1}{b x^{\frac{1}{3}}}\right)^{15}$ is equal to the coefficient of x^-15 in the expansion of $\left(\mathrm{ax}^{\frac{1}{3}}-\frac{1}{\mathrm{bx}^3}\right)^{15}$, where a and b are positive real numbers, then for each such ordered pair (a, b):

Accepted Answer

Since general term in the expansion of $\left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{15}$ is given by
¹⁵C_r(ax³)^15-r$\left(\frac{1}{b x^{1 / 3}}\right)^r$
Now 45 - 3r - $\frac r3$ = 15 $\Rightarrow$ 30 = $\frac {10r}{3}$ $\Rightarrow$ r = 9
So, coefficient of x¹⁵ = ¹⁵C₉a⁶b^-9
and general term in $\left(\mathrm{ax}^{1 / 3}-\frac{1}{\mathrm{bx}^3}\right)^{15}$
= ¹⁵C_r$\left(\mathrm{ax}^\frac {1}{ 3}\right)^{15-\mathrm{r}}\left(-\frac{1}{\mathrm{bx}^3}\right)^{\mathrm{r}}$
Now, 5 - $\frac r3$ - 3r = -15 $\Rightarrow$ $\frac {10r}{3}$ = 20 $\Rightarrow$ r = 6
So coefficient = ¹⁵C₆a⁹b^-6
Since, given $\frac{a^9}{b^6}=\frac{a^6}{b^9} \Rightarrow$ a³b³ = 1 $\Rightarrow$ ab = 1

Question:

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