Why Don't You Cube It First?

$\begin{aligned}&\sqrt[3] x+\sqrt[3] y=17\\\implies&x+y+3~\sqrt[3]{xy}(\sqrt[3]x+\sqrt[3] y)=17^3\\\implies&\sqrt[3]{xy}=\dfrac{17^3-(x+y)}{3(\sqrt[3]x+\sqrt[3] y)}\\\implies&xy=\left(\dfrac{17^3-(x+y)}{3(\sqrt[3]x+\sqrt[3] y)}\right)^3\\\implies&xy=\left(\dfrac{17^3-(3383)}{3(17)}\right)^3\\\implies&xy=(30)^3=\huge\boxed{27000}\end{aligned}$

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Formula used in second line: $\color{teal}{(x+y)^3=x^3+y^3+3xy(x+y)}$

Let's call $x = a^3, y = b^3 \Rightarrow a^3 + b^3 = 3383, \quad a + b = 17$ and then $a^3 + b^3 = 3383 = (a + b) \cdot (a^2 - ab + b^2) = 17 \cdot (a^2 - ab + b^2) \Rightarrow$ $a^2 - ab + b^2 = 199 \space \color{#D61F06}{(1)},\quad 17^2 = 289 = a^2 + 2ab + b^2 \space \color{#D61F06}{(2)}$ Substrating $\color{#D61F06}{(1)}$ to $\color{#D61F06}{(2)}$ we get $289 -199 = 90 = 3ab \Rightarrow ab = 30 \Rightarrow xy =a^3 \cdot b^3 = (ab)^{3} = 27000$

Rishabh, nice solution. But I simply made assumptions that to get 17... we need two no.s such that their no. Of digits is at least 3 in each case because when they add up , they give 3383. So , the possible combinations are 9,8 10,7 11,6 12,5 ..... and the last one 16,1. In all of these you find that 15 cube gives 3375 which is almost 3383. So, The Magical Combination is 15,2. Thus 15 cube i.e, 3375 and 2 cube i.e, 8 gives a product

3375×8=27000

$x+y=\big(\sqrt[3]x+\sqrt[3]y\big)\big(\sqrt[3]x^2-\sqrt[3]{xy}+\sqrt[3]y^2\big)=3383$

Substituting $\big(\sqrt[3]x+\sqrt[3]y\big)=17$

$17\big(\sqrt[3]x^2-\sqrt[3]{xy}+\sqrt[3]y^2\big)=3383$

And $\sqrt[3]x^2-\sqrt[3]{xy}+\sqrt[3]y^2=199$

Then, $\big(\sqrt[3]x+\sqrt[3]y\big)^2-\big(\sqrt[3]x^2-\sqrt[3]{xy}+\sqrt[3]y^2\big)=3\sqrt[3]{xy}=289-199=90$

Finally, $\sqrt[3]{xy}=30$ and $xy = 30^3 = \boxed{27000}$