what is the solution for the equation $\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}$?\n$b = - 4$ and $b =…

what is the solution for the equation $\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}$?\n$b = - 4$ and $b = 0$\n$b=-4$\n$b = 0$ and $b = 4$\n$b = 4$

what is the solution for the equation $\frac{5}{3b^{3}-2b^{2}-5}=\frac{2}{b^{3}-2}$?\n$b = - 4$ and $b = 0$\n$b=-4$\n$b = 0$ and $b = 4$\n$b = 4$

Answer

Explanation:

Step1: Cross - multiply

$5(b^{3}-2)=2(3b^{3}-2b^{2}-5)$

Step2: Expand both sides

$5b^{3}-10 = 6b^{3}-4b^{2}-10$

Step3: Move all terms to one side

$6b^{3}-4b^{2}-10-(5b^{3}-10)=0$ $6b^{3}-4b^{2}-10 - 5b^{3}+10=0$ $b^{3}-4b^{2}=0$

Step4: Factor out $b^{2}$

$b^{2}(b - 4)=0$

Step5: Set each factor equal to zero

$b^{2}=0$ gives $b = 0$; $b - 4=0$ gives $b=4$

Answer:

$b = 0$ and $b = 4$