Question

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

Answer 1

Option (c) is correct.

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

Explanation:

The given equation,
`(5)/(3b^3-3b^2-5)=(2)/(b^3-2)`

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

Consider,

`(5)/(3b^3-3b^2-5)=(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.