Find the value for this expression

Let $x = \sqrt{2} - 1$ . Then $a = bx$ , and the fraction in question becomes

$\dfrac{b^{3}x^{3} + 2b^{3}x^{2} + b^{3}}{b^{3}x(x + 3)} = \dfrac{x^{3} + 2x^{2} + 1}{x(x + 3)}$ ,

where $b^{3} \ne 0$ was cancelled out. This last expression can then be written as

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

Let $m$ be the solution of the expression given.

$\frac {a^3 + 2a^2 b + b^3}{ab(a+3b} = m$

Manipulating, we get

$a^3 + 2a^2 b + b^3 = mab(a+3b)$

$a^3 + 2a^2 b + b^3 + ab(a+3b) = mab(a+3b) + ab(a+3b)$

$a^3 + 3a^2 b + 3ab^2 + b^3 = (m+1)ab(a+b+2b)$

$(a + b)^3 = (m+1)ab(a+b+2b)$

From the given equation,

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

$a = b \sqrt 2 - b$

$a + b = b \sqrt 2 --------- \boxed {1}$

$(a + b)^3 = 2 \sqrt 2 b^3$

Equating both equations,

$(m+1)ab(a+b+2b) = 2 \sqrt 2 b^3$

$ab(m+1)(a+b+2b) = 2 \sqrt 2 b^3$

Substituting in $\boxed {1}$ ,

$ab(m+1)(b \sqrt 2+2b) = 2 \sqrt 2 b^3$

$ab^2(m+1)(\sqrt 2+2) = 2 \sqrt 2 b^3$

$a(m+1)(\sqrt 2+2) = 2 \sqrt 2 b$

$\frac {a}{b} \times (m+1) = \frac {2 \sqrt 2}{\sqrt 2+2}$

$\frac {a}{b} \times (m+1) = \frac {2 \sqrt 2}{\sqrt 2+2}$

$m+1 = \frac {2 \sqrt 2}{\sqrt 2+2} \div \frac {a}{b}$

Rationalizing the surd fraction,

$m+1 = \frac {2 \sqrt 2(\sqrt 2-2)}{(\sqrt 2+2)(\sqrt 2-2)} \div \frac {a}{b}$

$m+1 = \frac {2 \sqrt 2(\sqrt 2-2)}{2)} \div \frac {a}{b}$

$m+1 = \sqrt 2(\sqrt 2-2) \div \sqrt 2 -1$

$m+1 = 2 (\sqrt 2 -1) \div \sqrt 2 -1$

$m+1 = 2$

Therefore, $\boxed {m =1}$ .