Solve (3a + 1)/(9a ^ 2 + 3a + 1) + (3a - 1)/(9a ^ 2 - 3a + 1) - 2/(81a ^ 4 + 9a ^ 2 + 1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{3a+1}{9a^{2}+3a+1}+\frac{3a-1}{9a^{2}-3a+1}-\frac{2}{81a^{4}+9a^{2}+1}$$

Evaluate

$\frac{2\left(3a-1\right)}{9a^{2}-3a+1}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $9a^{2}+3a+1$ and $9a^{2}-3a+1$ is $\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)$. Multiply $\frac{3a+1}{9a^{2}+3a+1}$ times $\frac{9a^{2}-3a+1}{9a^{2}-3a+1}$. Multiply $\frac{3a-1}{9a^{2}-3a+1}$ times $\frac{9a^{2}+3a+1}{9a^{2}+3a+1}$.

$$\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}+\frac{\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Since $\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ and $\frac{\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ have the same denominator, add them by adding their numerators.

$$\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)+\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Do the multiplications in $\left(3a+1\right)\left(9a^{2}-3a+1\right)+\left(3a-1\right)\left(9a^{2}+3a+1\right)$.

$$\frac{27a^{3}-9a^{2}+3a+9a^{2}-3a+1+27a^{3}+9a^{2}+3a-9a^{2}-3a-1}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Combine like terms in $27a^{3}-9a^{2}+3a+9a^{2}-3a+1+27a^{3}+9a^{2}+3a-9a^{2}-3a-1$.

$$\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Factor $81a^{4}+9a^{2}+1$.

$$\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Since $\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ and $\frac{2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{54a^{3}-2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Factor the expressions that are not already factored in $\frac{54a^{3}-2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$.

$$\frac{2\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Cancel out $9a^{2}+3a+1$ in both numerator and denominator.

$$\frac{2\left(3a-1\right)}{9a^{2}-3a+1}$$

Use the distributive property to multiply $2$ by $3a-1$.

$$\frac{6a-2}{9a^{2}-3a+1}$$

Expand

$\frac{2\left(3a-1\right)}{9a^{2}-3a+1}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $9a^{2}+3a+1$ and $9a^{2}-3a+1$ is $\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)$. Multiply $\frac{3a+1}{9a^{2}+3a+1}$ times $\frac{9a^{2}-3a+1}{9a^{2}-3a+1}$. Multiply $\frac{3a-1}{9a^{2}-3a+1}$ times $\frac{9a^{2}+3a+1}{9a^{2}+3a+1}$.

$$\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}+\frac{\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Since $\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ and $\frac{\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ have the same denominator, add them by adding their numerators.

$$\frac{\left(3a+1\right)\left(9a^{2}-3a+1\right)+\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Do the multiplications in $\left(3a+1\right)\left(9a^{2}-3a+1\right)+\left(3a-1\right)\left(9a^{2}+3a+1\right)$.

$$\frac{27a^{3}-9a^{2}+3a+9a^{2}-3a+1+27a^{3}+9a^{2}+3a-9a^{2}-3a-1}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Combine like terms in $27a^{3}-9a^{2}+3a+9a^{2}-3a+1+27a^{3}+9a^{2}+3a-9a^{2}-3a-1$.

$$\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{81a^{4}+9a^{2}+1}$$

Factor $81a^{4}+9a^{2}+1$.

$$\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}-\frac{2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Since $\frac{54a^{3}}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ and $\frac{2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{54a^{3}-2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Factor the expressions that are not already factored in $\frac{54a^{3}-2}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$.

$$\frac{2\left(3a-1\right)\left(9a^{2}+3a+1\right)}{\left(9a^{2}-3a+1\right)\left(9a^{2}+3a+1\right)}$$

Cancel out $9a^{2}+3a+1$ in both numerator and denominator.

$$\frac{2\left(3a-1\right)}{9a^{2}-3a+1}$$

Use the distributive property to multiply $2$ by $3a-1$.

$$\frac{6a-2}{9a^{2}-3a+1}$$