Given that $x^{2018}y^{2017} = 1/2$ and $x^{2016}y^{2019} = 8$, the value of $x^2 + y^3$ is

We have two equations with large exponents:

$x^{2018}y^{2017} = \small \dfrac{1}{2}$ ... (1)

$x^{2016}y^{2019} = 8$ ... (2)

We need to find $x^2 + y^3$.

The key insight: When we have two equations with similar exponential terms, we can often find relationships between the variables by dividing one equation by the other.

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Let's divide equation (1) by equation (2):

$\small \dfrac{x^{2018}y^{2017}}{x^{2016}y^{2019}} = \small \dfrac{\small \dfrac{1}{2}}{8}$

This eliminates the large exponents and gives us a simpler relationship between $x$ and $y$.

$\small \dfrac{\small \dfrac{1}{2}}{8} = \small \dfrac{1}{2} \times \small \dfrac{1}{8} = \small \dfrac{1}{16}$

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For the left side, we use the rule: $\small \dfrac{a^m}{a^n} = a^{m-n}$

$\small \dfrac{x^{2018}y^{2017}}{x^{2016}y^{2019}} = x^{2018-2016} \times y^{2017-2019} = x^2 \times y^{-2} = \small \dfrac{x^2}{y^2}$

So we have:

$\small \dfrac{x^2}{y^2} = \small \dfrac{1}{16}$

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Taking the square root of both sides:

$\small \dfrac{x}{y} = \small \pm\dfrac{1}{4}$

This gives us: $x = \small \pm\dfrac{1}{4}y$

We keep the ± sign because $\sqrt{\small \dfrac{1}{16}} = \small \pm\dfrac{1}{4}$

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Substitute $x = \small \pm\dfrac{1}{4}y$ into equation (1):

$\left(\small \pm\dfrac{1}{4}y\right)^{2018} y^{2017} = \small \dfrac{1}{2}$

Since 2018 is even, $\left(\small \pm\dfrac{1}{4}\right)^{2018} = \left(\small \dfrac{1}{4}\right)^{2018}$

Any negative number raised to an even power becomes positive.

$\left(\small \dfrac{1}{4}\right)^{2018} y^{2018} y^{2017} = \small \dfrac{1}{2}$

$\left(\small \dfrac{1}{4}\right)^{2018} y^{4035} = \small \dfrac{1}{2}$

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$\small \dfrac{1}{4^{2018}} y^{4035} = \small \dfrac{1}{2}$

$y^{4035} = \small \dfrac{4^{2018}}{2}$

Since $4 = 2^2$, we have $4^{2018} = (2^2)^{2018} = 2^{4036}$

$y^{4035} = \small \dfrac{2^{4036}}{2} = 2^{4035}$

Therefore: $y = 2$

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Using $x = \small \pm\dfrac{1}{4}y$ and $y = 2$:

$x = \small \pm\dfrac{1}{4}(2) = \small \pm\dfrac{1}{2}$

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$x^2 = \left(\small \pm\dfrac{1}{2}\right)^2 = \small \dfrac{1}{4}$

$y^3 = 2^3 = 8$

$x^2 + y^3 = \small \dfrac{1}{4} + 8 = \small \dfrac{1}{4} + \small \dfrac{32}{4} = \small \dfrac{33}{4}$

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Answer: $\small \dfrac{33}{4}$