Question

what is the solution for the equation \\(\frac{5}{3b^3 - 2b^2 - 5} = \frac{2}{b^3 - 2}\\)?
options:
\\(b = -4\\)
\\(b = 0\\) and \\(b = 4\\)
\\(b = 4\\)
\\(b = -4\\) and \\(b = 0\\)

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}\), cross - multiplying (which is valid when \(3b^{3}-2b^{2}-5
eq0\) and \(b^{3}-2
eq0\)) gives us \(5(b^{3}-2)=2(3b^{3}-2b^{2}-5)\).

Step2: Expand both sides

Expand the left - hand side: \(5b^{3}-10\).
Expand the right - hand side: \(6b^{3}-4b^{2}-10\).
So the equation becomes \(5b^{3}-10 = 6b^{3}-4b^{2}-10\).

Step3: Simplify the equation

Subtract \(5b^{3}-10\) from both sides:
\(0=6b^{3}-4b^{2}-10-(5b^{3}-10)\)
\(0 = 6b^{3}-4b^{2}-10 - 5b^{3}+10\)
\(0=b^{3}-4b^{2}\)

Step4: Factor the equation

Factor out \(b^{2}\) from \(b^{3}-4b^{2}\): \(b^{2}(b - 4)=0\)

Step5: Solve for \(b\)

Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we have:
Case 1: \(b^{2}=0\) gives \(b = 0\). But we need to check if \(b = 0\) is a valid solution.
When \(b = 0\), the denominator of the left - hand side \(3b^{3}-2b^{2}-5=3(0)^{3}-2(0)^{2}-5=- 5
eq0\), and the denominator of the right - hand side \(b^{3}-2=0^{3}-2=-2
eq0\). But if we substitute \(b = 0\) into the original equation \(\frac{5}{3(0)^{3}-2(0)^{2}-5}=\frac{5}{-5}=-1\) and \(\frac{2}{0^{3}-2}=\frac{2}{-2}=-1\). Wait, but let's check the factoring again. Wait, maybe I made a mistake in cross - multiplying. Wait, the original equation: Let's re - examine the original equation. Wait, the user's equation: \(\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}\). Wait, when \(b = 0\), left denominator: \(3(0)-2(0)-5=-5\), right denominator: \(0 - 2=-2\). The equation becomes \(\frac{5}{-5}=\frac{2}{-2}\), i.e., \(-1=-1\), so \(b = 0\) is a solution? But when we factored, we got \(b^{2}(b - 4)=0\), so \(b = 0\) (double root) or \(b = 4\). Wait, but let's check \(b = 4\):
Left denominator: \(3(4)^{3}-2(4)^{2}-5=3\times64-2\times16 - 5=192 - 32-5 = 155
eq0\)
Right denominator: \(4^{3}-2=64 - 2=62
eq0\)
Substitute \(b = 4\) into the original equation: \(\frac{5}{3(4)^{3}-2(4)^{2}-5}=\frac{5}{192 - 32-5}=\frac{5}{155}=\frac{1}{31}\)
\(\frac{2}{4^{3}-2}=\frac{2}{62}=\frac{1}{31}\). So \(b = 4\) is a solution. And \(b = 0\): as we saw, \(\frac{5}{-5}=\frac{2}{-2}\), so \(b = 0\) is also a solution? But the options include \(b = 0\) and \(b = 4\). Wait, the options are: \(b=-4\), \(b = 0\) and \(b = 4\), \(b = 4\), \(b=-4\) and \(b = 0\). Wait, let's re - do the algebra.

Starting over:
\(\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}\)
Cross - multiply: \(5(b^{3}-2)=2(3b^{3}-2b^{2}-5)\)
\(5b^{3}-10 = 6b^{3}-4b^{2}-10\)
Subtract \(5b^{3}-10\) from both sides:
\(0=6b^{3}-4b^{2}-10 - 5b^{3}+10\)
\(0=b^{3}-4b^{2}\)
\(b^{2}(b - 4)=0\)
So \(b = 0\) or \(b = 4\). Now check \(b = 0\):
Left side: \(\frac{5}{3(0)^{3}-2(0)^{2}-5}=\frac{5}{-5}=-1\)
Right side: \(\frac{2}{0^{3}-2}=\frac{2}{-2}=-1\). So \(b = 0\) is a valid solution. And \(b = 4\) is also a valid solution. So the solution is \(b = 0\) and \(b = 4\).

Answer:

\(b = 0\) and \(b = 4\) (the option with \(b = 0\) and \(b = 4\))