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}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3b%5E3-3b%5E2-5%7D%3D%5Cfrac%7B2%7D%7Bb%5E3-2%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3b%5E3-3b%5E2-5%7D%3D%5Cfrac%7B2%7D%7Bb%5E3-2%7D)
Cross multiply ,
![5(b^3-2)=2(3b^3-2b^2-5)](https://tex.z-dn.net/?f=5%28b%5E3-2%29%3D2%283b%5E3-2b%5E2-5%29)
Multiply each term of LHS by 5 and RHS by 2 , we get,
![5b^3-10=6b^3-4b^2-10](https://tex.z-dn.net/?f=5b%5E3-10%3D6b%5E3-4b%5E2-10)
Taking variable terms one sides and constant one side, we get,
![5b^3-6b^3+4b^2=10-10](https://tex.z-dn.net/?f=5b%5E3-6b%5E3%2B4b%5E2%3D10-10)
Solving , we get,
![-b^3+4b^2=0](https://tex.z-dn.net/?f=-b%5E3%2B4b%5E2%3D0)
Taking
common from both terms, we get,
![b^2(4-b)=0](https://tex.z-dn.net/?f=b%5E2%284-b%29%3D0)
and ![\Rightarrow 4-b=0](https://tex.z-dn.net/?f=%5CRightarrow%204-b%3D0)
and ![\Rightarrow b=4](https://tex.z-dn.net/?f=%5CRightarrow%20b%3D4)
Thus, Option (c) is correct.
The solution of equation is b = 0 and b = 4.