Simplify (∛a + ∛b) divided by (2 + the cube root of a/b + the cube root of b/a).

Step-by-step solution

1. 1

   Substitute to hide the radicals. Let

   $x=\sqrt[3]{a},\qquad y=\sqrt[3]{b},$

   so that $\sqrt[3]{a/b}=x/y$ and $\sqrt[3]{b/a}=y/x$. The expression becomes purely rational in $x$ and $y$:

   $\frac{x+y}{2+\dfrac{x}{y}+\dfrac{y}{x}}.$

   This substitution is the point of the problem: cube roots that look unrelated are really just $x$ and $y$.

2. 2

   Put the denominator over $xy$.

   $2+\frac{x}{y}+\frac{y}{x}=\frac{2xy+x^{2}+y^{2}}{xy}.$

3. 3

   Recognise the numerator of that fraction as a square. Since $x^{2}+2xy+y^{2}=(x+y)^{2}$,

   $2+\frac{x}{y}+\frac{y}{x}=\frac{(x+y)^{2}}{xy}.$

   The same factor $x+y$ that sits on top of the original expression has now appeared squared underneath — that is what makes the cancellation possible.

4. 4

   Divide by inverting the fraction.

   $\frac{x+y}{\dfrac{(x+y)^{2}}{xy}}=(x+y)\cdot\frac{xy}{(x+y)^{2}}=\frac{xy}{x+y}.$

5. 5

   Translate back and check. With $xy=\sqrt[3]{a}\sqrt[3]{b}=\sqrt[3]{ab}$ and $x+y=\sqrt[3]{a}+\sqrt[3]{b}$,

   $\frac{\sqrt[3]{ab}}{\sqrt[3]{a}+\sqrt[3]{b}},$

   valid whenever $a\ne0$, $b\ne0$ and $\sqrt[3]{a}+\sqrt[3]{b}\ne0$. Testing $a=8$, $b=27$: $x=2$, $y=3$, the original denominator is $2+\tfrac23+\tfrac32=\tfrac{25}{6}$ and the numerator is $5$, giving $6/5$; the formula gives $xy/(x+y)=6/5$ as well.