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A-x-2ax-2-1-3-2a-2a-2-x-1-3-1-1-x-1-3-2a-1-3-2a-1-3-6-a-b-R-2-a-A-16a-4-x-2-b-A-8-ax-c-A-2a-1-3-3x




Question Number 146764 by Ar Brandon last updated on 15/Jul/21
A=((((x+((2ax^2 ))^(1/3) )(2a+((2a^2 x))^(1/3) )^(−1) −1)/( (x)^(1/3) −((2a))^(1/3) ))−(2a)^(−1/3) )^(−6) , (a,b)∈R^2   a- A=((16a^4 )/x^2 )             b- A=(8/(ax))             c-A=(((2a))^(1/3) /(3x^3 ))
$$\mathrm{A}=\left(\frac{\left(\mathrm{x}+\sqrt[{\mathrm{3}}]{\mathrm{2ax}^{\mathrm{2}} }\right)\left(\mathrm{2a}+\sqrt[{\mathrm{3}}]{\mathrm{2a}^{\mathrm{2}} \mathrm{x}}\right)^{−\mathrm{1}} −\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{2a}}}−\left(\mathrm{2a}\right)^{−\mathrm{1}/\mathrm{3}} \right)^{−\mathrm{6}} ,\:\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{a}-\:\mathrm{A}=\frac{\mathrm{16a}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}-\:\mathrm{A}=\frac{\mathrm{8}}{\mathrm{ax}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}-\mathrm{A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2a}}}{\mathrm{3x}^{\mathrm{3}} } \\ $$

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