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

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

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Answer 1 · 2014-11-01T12:45:39

$b^{2} - \frac{5}{3 b} = 0$ $b^2-5/{3b}=0$

by multiplying by $b$ $b$ ,

$\Rightarrow b^{3} - \frac{5}{3} = 0$ $=> b^3-5/3=0$

by $\frac{5}{3} = {(\sqrt[3]{\frac{5}{3}})}^{3}$ $5/3=(root3{5/3})^3$ ,

$\Rightarrow b^{3} - {(\sqrt[3]{\frac{5}{3}})}^{3} = 0$ $=> b^3-(root3{5/3})^3=0$

by $(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$ $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ ,

$\Rightarrow (b - \sqrt[3]{\frac{5}{3}}) [b^{2} + \sqrt[3]{\frac{5}{3}} b + {(\sqrt[3]{\frac{5}{3}})}^{2}] = 0$ $=> (b-root3{5/3})[b^2+root3{5/3}b+(root3{5/3})^2]=0$

since $b^{2} + \sqrt[3]{\frac{5}{3}} b + {(\sqrt[3]{\frac{5}{3}})}^{2} \neq 0$ $b^2+root3{5/3}b+(root3{5/3})^2 ne 0$ ,

$\Rightarrow b - \sqrt[3]{\frac{5}{3}} = 0 \Rightarrow b = \sqrt[3]{\frac{5}{3}}$ $=> b-root3{5/3}=0 => b=root3{5/3}$

I hope that this was helpful

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