Solve a^{2}*a^4+2a*a^5-3a^3)^2

$$a ^ { 2 } \times a ^ { 4 } + 2 a \times a ^ { 5 } - 3 a ^ { 3 } ) ^ { 2 }$$

Evaluate

$$3a^{3}\left(a^{3}-1\right)$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 4 to get 6.

$$a^{6}+2aa^{5}-3a^{3}$$

To multiply powers of the same base, add their exponents. Add 1 and 5 to get 6.

$$a^{6}+2a^{6}-3a^{3}$$

Combine a^{6} and 2a^{6} to get 3a^{6}.

$$3a^{6}-3a^{3}$$

Factor

$$3\left(a-1\right)\left(a^{2}+a+1\right)a^{3}$$

Solution Steps

Factor out a.

$$a\left(aa^{4}+2a^{5}-3a^{2}\right)$$

Consider a^{5}+2a^{5}-3a^{2}. Factor out a^{2}.

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

Consider a^{3}+2a^{3}-3. Multiply and combine like terms.

$$3a^{3}-3$$

Consider 3a^{3}-3. Factor out 3.

$$3\left(a^{3}-1\right)$$

Consider a^{3}-1. Rewrite a^{3}-1 as a^{3}-1^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

$$\left(a-1\right)\left(a^{2}+a+1\right)$$

Rewrite the complete factored expression. Polynomial a^{2}+a+1 is not factored since it does not have any rational roots.

$$3aa^{2}\left(a-1\right)\left(a^{2}+a+1\right)$$