Solve (4x^2-9)^3

$$( 4 x ^ { 2 } - 9 ) ^ { 3 }$$

Expand

$$64x^{6}-432x^{4}+972x^{2}-729$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4x^{2}-9\right)^{3}.

$$64\left(x^{2}\right)^{3}-432\left(x^{2}\right)^{2}+972x^{2}-729$$

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

$$64x^{6}-432\left(x^{2}\right)^{2}+972x^{2}-729$$

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

$$64x^{6}-432x^{4}+972x^{2}-729$$

Evaluate

$$\left(4x^{2}-9\right)^{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4x^{2}-9\right)^{3}.

$$64\left(x^{2}\right)^{3}-432\left(x^{2}\right)^{2}+972x^{2}-729$$

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

$$64x^{6}-432\left(x^{2}\right)^{2}+972x^{2}-729$$

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

$$64x^{6}-432x^{4}+972x^{2}-729$$