Solve 1/(3a-1)(9a^2+3a+1):1+3a/1+3a+9a^2

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

Evaluate

$$\frac{1}{9a^{2}-1}$$

Solution Steps

Divide \frac{1}{\left(3a-1\right)\left(9a^{2}+3a+1\right)} by \frac{1+3a}{1+3a+9a^{2}} by multiplying \frac{1}{\left(3a-1\right)\left(9a^{2}+3a+1\right)} by the reciprocal of \frac{1+3a}{1+3a+9a^{2}}.

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

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

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

Consider \left(3a-1\right)\left(3a+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(3a\right)^{2}.

$$\frac{1}{3^{2}a^{2}-1}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{9a^{2}-1}$$

Expand

$$\frac{1}{9a^{2}-1}$$

Solution Steps

Divide \frac{1}{\left(3a-1\right)\left(9a^{2}+3a+1\right)} by \frac{1+3a}{1+3a+9a^{2}} by multiplying \frac{1}{\left(3a-1\right)\left(9a^{2}+3a+1\right)} by the reciprocal of \frac{1+3a}{1+3a+9a^{2}}.

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

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

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

Consider \left(3a-1\right)\left(3a+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(3a\right)^{2}.

$$\frac{1}{3^{2}a^{2}-1}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{9a^{2}-1}$$