If a^3 + b^6 = 2,then the maximum value of th | vedclass

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(C) The general term $T_{r+1}$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is given by $T_{r+1} = {^9C_r} (ax^{1/3})^{9-r} (bx^{-1/6})^r$.For the t

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$84$

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(C) The general term $T_{r+1}$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is given by $T_{r+1} = {^9C_r} (ax^{1/3})^{9-r} (bx^{-1/6})^r$.For the term to be independent of $x$,the power of $x$ must be zero:$\frac{9-r}{3} - \frac{r}{6} = 0$ $\Rightarrow \frac{18-2r-r}{6} = 0$ $\Rightarrow 18-3r=0$ $\Rightarrow r=6$.The independent term is $T_{6+1} = {^9C_6} a^3 b^6 = 84 a^3 b^6$.Given $a^3 + b^6 = 2$,by the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality:$\frac{a^3 + b^6}{2} \geq \sqrt{a^3 b^6}$ $\Rightarrow \frac{2}{2} \geq \sqrt{a^3 b^6}$ $\Rightarrow 1 \geq \sqrt{a^3 b^6}$ $\Rightarrow a^3 b^6 \leq 1$.Therefore,the maximum value of the independent term is $84 	imes 1 = 84$.

If $a^3 + b^6 = 2$,then the maximum value of the term independent of $x$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is,where $(a > 0, b > 0)$.

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