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

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Question

$r=\frac{3}{\frac{1}{3}+\frac{1}{6}}=6$

$Tr + 1 = T6 + 1 = {\,^9}{c_6}{a^3}{b^6} = 84{a^3}{b^6}$

$A \geq G$

$\frac{a^{3}+b^{6}}{2} \geq \sqrt

Vedclass · Accepted Answer

$84$

Vedclass · Answer

$r=\frac{3}{\frac{1}{3}+\frac{1}{6}}=6$

$Tr + 1 = T6 + 1 = {\,^9}{c_6}{a^3}{b^6} = 84{a^3}{b^6}$

$A \geq G$

$\frac{a^{3}+b^{6}}{2} \geq \sqrt{a^{3} b^{6}}$

$1 \geq a^{3} b^{6} \Rightarrow a^{3} b^{6} \leq 1$

So $84 a^{3} b^{6} \leq 84$

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

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