Answer to Question #191706 in Algebra for Abundance

Question

Find the value of a³+2a²b+b³ divided by ab(a+3b) when a/b=√2-1

Answer 1

We have given that,

$\dfrac{a}{b} = \sqrt{2}-1$

$a = b(\sqrt{2}-1)$

We have to find the value of $\dfrac{a^3+2a^b+b^3}{ab(a+3b)}$

Putting the value of a in the above equation,

We get,

$\dfrac{b^3(\sqrt{2}-1)^3+2b^3(\sqrt{2}-1)^2+b^3}{b^3(\sqrt{2}-1)^2+3b^3(\sqrt{2}-1)}$

$= \dfrac{(\sqrt{2}-1)^3+(\sqrt{2}-1)^2+1}{(\sqrt{2}-1)^2+(\sqrt{2}-1)}$

After solving we get,

$\dfrac{3 \sqrt{2}-1}{2-\sqrt{2}}$

$= \dfrac{3}{\sqrt{2}}$