If the coefficient of x^{15} in the expansion | vedclass

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(B) For the expansion $(ax^3 + \frac{1}{bx^{1/3}})^{15}$,the general term is $T_{r+1} = {}^{15}C_r (ax^3)^{15-r} (b^{-1}x^{-1/3})^r = {}^{15}C_r a^{15

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$ab=1$

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(B) For the expansion $(ax^3 + \frac{1}{bx^{1/3}})^{15}$,the general term is $T_{r+1} = {}^{15}C_r (ax^3)^{15-r} (b^{-1}x^{-1/3})^r = {}^{15}C_r a^{15-r} b^{-r} x^{45-3r-r/3}$.Setting the exponent of $x$ to $15$: $45 - \frac{10r}{3} = 15$ $\Rightarrow \frac{10r}{3} = 30$ $\Rightarrow r = 9$.The coefficient is ${}^{15}C_9 a^6 b^{-9}$.For the expansion $(ax^{1/3} - \frac{1}{bx^3})^{15}$,the general term is $T_{r+1} = {}^{15}C_r (ax^{1/3})^{15-r} (-b^{-1}x^{-3})^r = {}^{15}C_r a^{15-r} (-1)^r b^{-r} x^{5-r/3-3r}$.Setting the exponent of $x$ to $-15$: $5 - \frac{10r}{3} = -15$ $\Rightarrow \frac{10r}{3} = 20$ $\Rightarrow r = 6$.The coefficient is ${}^{15}C_6 a^9 (-1)^6 b^{-6} = {}^{15}C_6 a^9 b^{-6}$.Since ${}^{15}C_9 = {}^{15}C_6$,we equate the coefficients: $a^6 b^{-9} = a^9 b^{-6}$.Dividing both sides by $a^6 b^{-6}$,we get $b^{-3} = a^3$,which implies $a^3 b^3 = 1$,so $ab = 1$.

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

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