Respuesta :
Answer:
The solutions are:
[tex]b=0,\:b=4[/tex]
Step-by-step explanation:
Considering the expression
- [tex]\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}[/tex]
Solving the expression
[tex]\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}[/tex]
[tex]\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c[/tex]
[tex]5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2[/tex]
[tex]5b^3-10=6b^3-4b^2-10[/tex]
[tex]\mathrm{Switch\:sides}[/tex]
[tex]6b^3-4b^2-10=5b^3-10[/tex]
[tex]6b^3-4b^2-10+10=5b^3-10+10[/tex]
[tex]6b^3-4b^2=5b^3[/tex]
[tex]\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}[/tex]
[tex]6b^3-4b^2-5b^3=5b^3-5b^3[/tex]
[tex]b^3-4b^2=0[/tex]
[tex]Using\:the\:Zero\:Factor\:Principle:[/tex] [tex]if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
So,
[tex]b=0,b-4=0[/tex]
[tex]b=0,b=4[/tex]
Therefore, the solutions are:
[tex]b=0,\:b=4[/tex]