Solve (1/9n^2n+4)(1/9n^2n-4)

$$( \frac { 1 } { 9 } n ^ { 2 } n + 4 ) ( \frac { 1 } { 9 } n ^ { 2 } n - 4 )$$

Evaluate

$$\frac{n^{6}}{81}-16$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$\left(\frac{1}{9}n^{3}\right)^{2}-16$$

Expand \left(\frac{1}{9}n^{3}\right)^{2}.

$$\left(\frac{1}{9}\right)^{2}\left(n^{3}\right)^{2}-16$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\left(\frac{1}{9}\right)^{2}n^{6}-16$$

Calculate \frac{1}{9} to the power of 2 and get \frac{1}{81}.

$$\frac{1}{81}n^{6}-16$$

Expand

$$\frac{n^{6}}{81}-16$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$\left(\frac{1}{9}n^{3}\right)^{2}-16$$

Expand \left(\frac{1}{9}n^{3}\right)^{2}.

$$\left(\frac{1}{9}\right)^{2}\left(n^{3}\right)^{2}-16$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\left(\frac{1}{9}\right)^{2}n^{6}-16$$

Calculate \frac{1}{9} to the power of 2 and get \frac{1}{81}.

$$\frac{1}{81}n^{6}-16$$