Solve for b 5/(3b^3-2b^2-5)=2/(b^3-2)

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

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$5 (b^{3} - 2) = (3 b^{3} - 2 b^{2} - 5) \cdot 2$

Step 2

Solve the equation for $b$ .

Since $b$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$(3 b^{3} - 2 b^{2} - 5) \cdot 2 = 5 (b^{3} - 2)$

Simplify $(3 b^{3} - 2 b^{2} - 5) \cdot 2$ .

Rewrite.

$0 + 0 + (3 b^{3} - 2 b^{2} - 5) \cdot 2 = 5 (b^{3} - 2)$

Simplify by adding zeros.

$(3 b^{3} - 2 b^{2} - 5) \cdot 2 = 5 (b^{3} - 2)$

Apply the distributive property.

$3 b^{3} \cdot 2 - 2 b^{2} \cdot 2 - 5 \cdot 2 = 5 (b^{3} - 2)$

Simplify.

Multiply $2$ by $3$ .

$6 b^{3} - 2 b^{2} \cdot 2 - 5 \cdot 2 = 5 (b^{3} - 2)$

Multiply $2$ by $- 2$ .

$6 b^{3} - 4 b^{2} - 5 \cdot 2 = 5 (b^{3} - 2)$

Multiply $- 5$ by $2$ .

$6 b^{3} - 4 b^{2} - 10 = 5 (b^{3} - 2)$

Simplify $5 (b^{3} - 2)$ .

Apply the distributive property.

$6 b^{3} - 4 b^{2} - 10 = 5 b^{3} + 5 \cdot - 2$

Multiply $5$ by $- 2$ .

$6 b^{3} - 4 b^{2} - 10 = 5 b^{3} - 10$

Move all terms containing $b$ to the left side of the equation.

Subtract $5 b^{3}$ from both sides of the equation.

$6 b^{3} - 4 b^{2} - 10 - 5 b^{3} = - 10$

Subtract $5 b^{3}$ from $6 b^{3}$ .

$b^{3} - 4 b^{2} - 10 = - 10$

Add $10$ to both sides of the equation.

$b^{3} - 4 b^{2} - 10 + 10 = 0$

Combine the opposite terms in $b^{3} - 4 b^{2} - 10 + 10$ .

Add $- 10$ and $10$ .

$b^{3} - 4 b^{2} + 0 = 0$

Add $b^{3} - 4 b^{2}$ and $0$ .

$b^{3} - 4 b^{2} = 0$

Factor $b^{2}$ out of $b^{3} - 4 b^{2}$ .

Factor $b^{2}$ out of $b^{3}$ .

$b^{2} b - 4 b^{2} = 0$

Factor $b^{2}$ out of $- 4 b^{2}$ .

$b^{2} b + b^{2} \cdot - 4 = 0$

Factor $b^{2}$ out of $b^{2} b + b^{2} \cdot - 4$ .

$b^{2} (b - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$b^{2} = 0$

$b - 4 = 0$

Set $b^{2}$ equal to $0$ and solve for $b$ .

Set $b^{2}$ equal to $0$ .

$b^{2} = 0$

Solve $b^{2} = 0$ for $b$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$b = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 0$

Plus or minus $0$ is $0$ .

$b = 0$

Set $b - 4$ equal to $0$ and solve for $b$ .

Set $b - 4$ equal to $0$ .

$b - 4 = 0$

Add $4$ to both sides of the equation.

$b = 4$

The final solution is all the values that make $b^{2} (b - 4) = 0$ true.

$b = 0, 4$

Solve for b 5/(3b^3-2b^2-5)=2/(b^3-2)

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