Solve (9a^4-1/4b^2)*(9a^4+1/4b^2)

$$( 9 a ^ { 4 } - \frac { 1 } { 4 } b ^ { 2 } ) \times ( 9 a ^ { 4 } + \frac { 1 } { 4 } b ^ { 2 } )$$

Evaluate

$$81a^{8}-\frac{b^{4}}{16}$$

Solution Steps

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

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

Expand \left(9a^{4}\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$9^{2}a^{8}-\left(\frac{1}{4}b^{2}\right)^{2}$$

Calculate 9 to the power of 2 and get 81.

$$81a^{8}-\left(\frac{1}{4}b^{2}\right)^{2}$$

Expand \left(\frac{1}{4}b^{2}\right)^{2}.

$$81a^{8}-\left(\frac{1}{4}\right)^{2}\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$81a^{8}-\left(\frac{1}{4}\right)^{2}b^{4}$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$81a^{8}-\frac{1}{16}b^{4}$$

Expand

$$81a^{8}-\frac{b^{4}}{16}$$

Solution Steps

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

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

Expand \left(9a^{4}\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$9^{2}a^{8}-\left(\frac{1}{4}b^{2}\right)^{2}$$

Calculate 9 to the power of 2 and get 81.

$$81a^{8}-\left(\frac{1}{4}b^{2}\right)^{2}$$

Expand \left(\frac{1}{4}b^{2}\right)^{2}.

$$81a^{8}-\left(\frac{1}{4}\right)^{2}\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$81a^{8}-\left(\frac{1}{4}\right)^{2}b^{4}$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$81a^{8}-\frac{1}{16}b^{4}$$