Question

Help plz................

Answer 1

Answer:

Option (c) is correct.

The solution of equation is b = 0 and b = 4.

Step-by-step explanation:

 The given equation, $\frac{5}{3b^3-3b^2-5}=\frac{2}{b^3-2}$

We are required to solve the given equation for possible values of b.

Consider,

$\frac{5}{3b^3-3b^2-5}=\frac{2}{b^3-2}$

Cross multiply ,

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

Multiply each term of LHS by 5 and RHS by 2 , we get,

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

Taking variable terms one sides and constant one side, we get,

$5b^3-6b^3+4b^2=10-10$

Solving , we get,

$-b^3+4b^2=0$

Taking $b^2$ common from both terms, we get,

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

$\Rightarrow b^2=0$ and $\Rightarrow 4-b=0$

$\Rightarrow b=0$ and  $\Rightarrow b=4$

Thus, Option (c) is correct.

The solution of equation is b = 0 and b = 4.