Find the value of the expression $A=xy^3-x^3y$

Question

Let $$x=\frac{2}{2\cdot \:2^{\frac{1}{3}}+2+4^{\frac{1}{3}}};y=\frac{2}{2\cdot \:2^{\frac{1}{3}}-2+4^{\frac{1}{3}}}$$.Find the value of the expression $$A=xy^3-x^3y$$

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I see: $$\left(4^{\frac{1}{3}}-2^{\frac{1}{3}}\right)\left(4^{\frac{2}{3}}+\left(4\cdot 2\right)^{\frac{1}{3}}+2^{\frac{2}{3}}\right)=4-2=2$$

$$\left(4^{\frac{1}{3}}+2^{\frac{1}{3}}\right)\left(4^{\frac{2}{3}}-\left(4\cdot \:2\right)^{\frac{1}{3}}+2^{\frac{2}{3}}\right)=4+2=6$$

So $A=xy\left(x+y\right)\left(y-x\right)$

And i found $A=\frac{-2\left(16^{\frac{1}{3}}-4^{\frac{1}{3}}\right)^2}{9}+\frac{-4\cdot 4^{\frac{1}{3}}\left(16^{\frac{1}{3}}-4^{\frac{1}{3}}\right)\left(4^{\frac{1}{3}}+2^{\frac{1}{3}}\right)}{27}$ but that's very bad. Help me improve it

Answer 1

Let $a=2^{\dfrac{1}{3}}$ therefore $$x=\dfrac{a}{a^2+a+1}\\y=\dfrac{a}{a^2-a+1}$$therefore$$A{=xy(y^2-x^2)\\=xy(y+x)(y-x)\\=\dfrac{a^2}{a^4+a^2+1}\dfrac{2a^3+2a}{a^4+a^2+1}\dfrac{2a^2}{a^4+a^2+1}\\=\dfrac{8a^2(a^2+1)}{(a+1)^6}\\=\dfrac{8a^2(a^2+1)}{(a^3+3a^2+3a+1)^2}\\=\dfrac{8a^2(a^2+1)}{9(a^2+a+1)^2}}$$since $(a-1)(a^2+1+a)=a^3-1=1$ we have $$A=\dfrac{8}{9}a^2(a^2+1)(a-1)^2=\dfrac{8}{9}a^2(a^4-2a^3+a^2+a^2-2a+1)=\dfrac{8}{9}a^2(2a^2-3)=\dfrac{8}{9}(4a-3a^2)$$

Answer 2

\begin{align} A &= \frac{-2\left(16^{\frac{1}{3}}-4^{\frac{1}{3}}\right)^2}{9}+\frac{-4\cdot 4^{\frac{1}{3}}\left(16^{\frac{1}{3}}-4^{\frac{1}{3}}\right)\left(4^{\frac{1}{3}}+2^{\frac{1}{3}}\right)}{27}\\ &= \frac{-2\left(2^{\frac{3}{3}}-2^{\frac{2}{3}}\right)^2}{9}+\frac{- 2^{\frac{8}{3}}\left(2^{\frac{3}{3}}-2^{\frac{2}{3}}\right)\left(2^{\frac{2}{3}}+2^{\frac{1}{3}}\right)}{27}\\ &= \frac{-2\left(2-2^{\frac{2}{3}}\right)^2}{9} +\frac{- 8\left(2^{\frac{2}{3}}-2^{\frac{1}{3}}\right)\left(2^{\frac{2}{3}}+2^{\frac{1}{3}}\right)}{27}\\ &= \frac{-2\left(2-2^{\frac{2}{3}}\right)^2}{9} +\frac{- 8\left(2^{\frac{4}{3}}-2^{\frac{2}{3}}\right)}{27}\\ & \qquad \vdots \end{align}

Answer 3

A less ugly way, but probably still not the best way:

$\dfrac{1}{x}=\dfrac{2\cdot \:2^{\frac{1}{3}}+2+4^{\frac{1}{3}}}{2}; \dfrac{1}{y}=\dfrac{2\cdot \:2^{\frac{1}{3}}-2+4^{\frac{1}{3}}}{2} \Rightarrow \dfrac{1}{x}-\dfrac{1}{y}=2\Rightarrow y-x=2xy$

$A=xy^3-x^3y$

$=xy(y^2-x^2)$

$=xy(y-x)(x+y)$

$=2x^2y^2(x+y)$

Next step:

$x=\dfrac{2}{2\cdot \:2^{\frac{1}{3}}+2+4^{\frac{1}{3}}}=\dfrac{2}{2^{4/3}+2+2^{2/3}}$

$y=\dfrac{2}{2\cdot \:2^{\frac{1}{3}}-2+4^{\frac{1}{3}}}=\dfrac{2}{2^{4/3}-2+2^{2/3}}$

$xy=\dfrac{4}{(2^{4/3}+2+4^{1/3})(2^{4/3}-2+2^{2/3})}=\dfrac{4}{(2^{4/3}+2^{2/3})^2-2^2}=\dfrac{4}{2^{8/3}+2^{4/3}+8-4}$

$x+y=\dfrac{2}{2^{4/3}+2+2^{2/3}}+\dfrac{2}{2^{4/3}-2+2^{2/3}}=\dfrac{2^{7/3}-4+2^{5/3}+2^{7/3}+4+2^{5/3}}{(2^{4/3}+2+4^{1/3})(2^{4/3}-2+2^{2/3})}$

$\Rightarrow {\begin{cases}xy=\dfrac{4}{2^{8/3}+2^{4/3}+4}\\x+y=\dfrac{2^{10/3}+2^{8/3}}{2^{8/3}+2^{4/3}+4}\end{cases}}$

$\Rightarrow 2x^2y^2(x+y)=\dfrac{32(2^{10/3}+2^{8/3})}{(2^{8/3}+2^{4/3}+4)^3}$

You can continue from here, unfortunately it is not a good number:

$0.2466498166...=\dfrac{8}{9} \sqrt[3]{2} \left(4-3 \sqrt[3]{2}\right)$

(I check with my calculator, this is the correct solution)