If x $\dfrac{\sqrt{2}-1}{\sqrt{2}+1}$ and y = $\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$ , then find the value of ${x^2 +5xy + y^2}$.

$x = \dfrac{\sqrt{2}-1}{\sqrt{2}+1} \text{ and y} = \dfrac{\sqrt{2}+1}{\sqrt{2}-1} \\[1.5em] \therefore x+y = \dfrac{\sqrt{2}-1}{\sqrt{2}+1} +\dfrac{\sqrt{2}+1}{\sqrt{2}-1} \\[1.5em] \Rightarrow\dfrac{(\sqrt{2}-1)^2 + (\sqrt{2}+1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)} \\[1.5em] \Rightarrow\dfrac{(\sqrt{2})^2 - 2\sqrt{2} + 1 + (\sqrt{2})^2 + 2\sqrt{2} + 1}{(\sqrt{2})^2 - 1} \\[1.5em] \Rightarrow\dfrac{2+1+2+1}{2-1} = 6 \\[1.5em] \Rightarrow x+y = 6 \qquad \text{….(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{2}-1}{\sqrt{2}+1} ×\dfrac{\sqrt{2}+1}{\sqrt{2}-1} = 1 \qquad \text{….(ii)} \\[1.5em]$

We need to find the value of ${x^2 +5xy + y^2}$ ${x^2 + 5xy + y^2} = x^2 + y^2 + 2xy + 3xy \\[1.5em] \Rightarrow {x^2 + 5xy + y^2} = (x+y)^2 + 3xy \qquad \text{….(iii)} \\[1.5em]$

substituting the values from (i) and (ii) in (iii),

${x^2 + 5xy + y^2} = (6)^2 + 3 × 1 \\[1.5em] = \bold{39}$