How do you factor and solve #b^2-\frac{5}{3b}=0#?

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Answer 1 · 2014-11-01T12:45:39

#b^2-5/{3b}=0#

by multiplying by #b#,

#=> b^3-5/3=0#

by #5/3=(root3{5/3})^3#,

#=> b^3-(root3{5/3})^3=0#

by #(a^3-b^3)=(a-b)(a^2+ab+b^2)#,

#=> (b-root3{5/3})[b^2+root3{5/3}b+(root3{5/3})^2]=0#

since #b^2+root3{5/3}b+(root3{5/3})^2 ne 0#,

#=> b-root3{5/3}=0 => b=root3{5/3}#

I hope that this was helpful