Solve (2-4n^3)^3

$$( 2 - 4 n ^ { 3 } ) ^ { 3 }$$

Evaluate

$$8\left(1-2n^{3}\right)^{3}$$

Solution Steps

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

$$8-48n^{3}+96\left(n^{3}\right)^{2}-64\left(n^{3}\right)^{3}$$

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

$$8-48n^{3}+96n^{6}-64\left(n^{3}\right)^{3}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$8-48n^{3}+96n^{6}-64n^{9}$$

Expand

$$8-48n^{3}+96n^{6}-64n^{9}$$

Solution Steps

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

$$8-48n^{3}+96\left(n^{3}\right)^{2}-64\left(n^{3}\right)^{3}$$

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

$$8-48n^{3}+96n^{6}-64\left(n^{3}\right)^{3}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$8-48n^{3}+96n^{6}-64n^{9}$$