Solve (a^2+3)^3

$$( a ^ { 2 } + 3 ) ^ { 3 }$$

Expand

$$a^{6}+9a^{4}+27a^{2}+27$$

Solution Steps

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

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

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

$$a^{6}+9\left(a^{2}\right)^{2}+27a^{2}+27$$

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

$$a^{6}+9a^{4}+27a^{2}+27$$

Evaluate

$$\left(a^{2}+3\right)^{3}$$

Solution Steps

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

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

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

$$a^{6}+9\left(a^{2}\right)^{2}+27a^{2}+27$$

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

$$a^{6}+9a^{4}+27a^{2}+27$$