## Task B-2.6.

For which integers $b$ is the expression

$$
\left(\frac{b}{b+8}-\frac{4 b}{\left(b^{\frac{1}{3}}+2\right)^{3}}\right) \cdot\left(\frac{1+2 b^{-\frac{1}{3}}}{1-2 b^{-\frac{1}{3}}}\right)^{2}+\frac{24}{b+8}
$$

greater than 1?
## Solution.

Simplify the given expression.

$$
\begin{aligned}
& \left(\frac{b}{b+8}-\frac{4 b}{\left(b^{\frac{1}{3}}+2\right)^{3}}\right) \cdot\left(\frac{1+2 b^{-\frac{1}{3}}}{1-2 b^{-\frac{1}{3}}}\right)^{2}+\frac{24}{b+8} \\
& =\left(\frac{b}{\left(b^{\frac{1}{3}}+2\right)\left(b^{\frac{2}{3}}-2 b^{\frac{1}{3}}+4\right)}-\frac{4 b}{\left(b^{\frac{1}{3}}+2\right)^{3}}\right) \cdot\left(\frac{\frac{b^{\frac{1}{3}}+2}{b^{\frac{1}{3}}}}{\frac{b^{\frac{1}{3}}-2}{b^{\frac{1}{3}}}}\right)^{2}+\frac{24}{b+8}
\end{aligned}
$$

1 point for factoring the first denominator, 1 point for addition in the second parenthesis

$$
=\frac{b\left(b^{\frac{1}{3}}+2\right)^{2}-4 b\left(b^{\frac{2}{3}}-2 b^{\frac{1}{3}}+4\right)}{\left(b^{\frac{1}{3}}+2\right)^{3}\left(b^{\frac{2}{3}}-2 b^{\frac{1}{3}}+4\right)} \cdot\left(\frac{b^{\frac{1}{3}}+2}{b^{\frac{1}{3}}-2}\right)^{2}+\frac{24}{b+8}
$$

reducing to a common denominator

$$
=\frac{b\left(b^{\frac{2}{3}}+4 b^{\frac{1}{3}}+4-4 b^{\frac{2}{3}}+8 b^{\frac{1}{3}}-16\right)}{\left(b^{\frac{1}{3}}+2\right)^{3}\left(b^{\frac{2}{3}}-2 b^{\frac{1}{3}}+4\right)} \cdot \frac{\left(b^{\frac{1}{3}}+2\right)^{2}}{\left(b^{\frac{1}{3}}-2\right)^{2}}+\frac{24}{b+8}
$$

factoring out $b$ and squaring in the numerator after reducing to a common denominator

$$
=\frac{b\left(-3 b^{\frac{2}{3}}+12 b^{\frac{1}{3}}-12\right)}{(b+8)\left(b^{\frac{1}{3}}-2\right)^{2}}+\frac{24}{b+8}
$$

simplifying the numerator and multiplying by the other fraction

$$
=\frac{-3 b\left(b^{\frac{1}{3}}-2\right)^{2}}{(b+8)\left(b^{\frac{1}{3}}-2\right)^{2}}+\frac{24}{b+8}
$$

noting the square of a binomial

$$
=\frac{-3 b}{b+8}+\frac{24}{b+8}=\frac{-3(b-8)}{b+8}
$$

simplifying and writing the final simplified expression

Solve the inequality $\frac{-3(b-8)}{b+8}>1$ in the set $\mathbb{Z}$.

$\frac{-3(b-8)}{b+8}>1 \Longleftrightarrow \frac{-3 b+24-b-8}{b+8}>0 \Longleftrightarrow \frac{-4 b+16}{b+8}>0 \Longleftrightarrow \frac{-b+4}{b+8}>0 \quad 1$ point

By considering cases or using a sign chart, we get the solution $b \in\langle-8,4\rangle$.

1 point

The final solution consists of all integers in this interval except zero, since for $b=0$ the original expression is undefined (a negative exponent appears).

Thus, $b \in\{-7,-6,-5,-4,-3,-2,-1,1,2,3\}$.

1 point

Note: If the student includes 0 in the solution, deduct 1 point.

Note: The problem can also be solved using the substitution $b=a^{3}$. Then we get the expression:

$$
\left(\frac{a^{3}}{a^{3}+8}-\frac{4 a^{3}}{(a+2)^{3}}\right) \cdot\left(\frac{1+2 a^{-1}}{1-2 a^{-1}}\right)^{2}+\frac{24}{a^{3}+8}
$$

and solve it as in the first method. The scoring is analogous.