Answer:
b = 0 and b = 4
Step-by-step explanation:
[tex]\frac{5}{3b^3-2b^2-5}[/tex] = [tex]\frac{2}{b^3-2}[/tex] ( cross- multiply )
2(3b³ - 2b² - 5) = 5(b³ - 2) ← distribute parenthesis on both sides
6b³ - 4b² - 10 = 5b³ - 10 ( subtract 5b³ from both sides )
b³ - 4b² - 10 = - 10 ( add 10 to both sides )
b³ - 4b² = 0 ← factor out b² from each term on the left side
b²(b - 4) = 0
equate each factor to zero and solve for b
b² = 0 ⇒ b = 0
b - 4 = 0 ⇒ b = 4
