Solve 4(a^3-3)^3

$$4 ( a ^ { 3 } - 3 ) ^ { 3 }$$

Expand

$$4a^{9}-36a^{6}+108a^{3}-108$$

Solution Steps

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a^{3}-3\right)^{3}.

$$4\left(\left(a^{3}\right)^{3}-9\left(a^{3}\right)^{2}+27a^{3}-27\right)$$

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

$$4\left(a^{9}-9\left(a^{3}\right)^{2}+27a^{3}-27\right)$$

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

$$4\left(a^{9}-9a^{6}+27a^{3}-27\right)$$

Use the distributive property to multiply 4 by a^{9}-9a^{6}+27a^{3}-27.

$$4a^{9}-36a^{6}+108a^{3}-108$$

Evaluate

$$4\left(a^{3}-3\right)^{3}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a^{3}-3\right)^{3}.

$$4\left(\left(a^{3}\right)^{3}-9\left(a^{3}\right)^{2}+27a^{3}-27\right)$$

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

$$4\left(a^{9}-9\left(a^{3}\right)^{2}+27a^{3}-27\right)$$

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

$$4\left(a^{9}-9a^{6}+27a^{3}-27\right)$$

Use the distributive property to multiply 4 by a^{9}-9a^{6}+27a^{3}-27.

$$4a^{9}-36a^{6}+108a^{3}-108$$