Solve (3^{-3}a^{3}+3^{-2}a^{2}+3^{1}a+3^0)(-9a^2)+a^2(a^2+27a+9)+1.overline{3}a

$$( 3 ^ { - 3 } a ^ { 3 } + 3 ^ { - 2 } a ^ { 2 } + 3 ^ { 1 } a + 3 ^ { 0 } ) ( - 9 a ^ { 2 } ) + a ^ { 2 } ( a ^ { 2 } + 27 a + 9 ) + 1 . \overline { 3 } a$$

Evaluate

$$-\frac{a^{5}}{3}+1$$

Solution Steps

Calculate 3 to the power of -3 and get \frac{1}{27}.

$$\left(\frac{1}{27}a^{3}+3^{-2}a^{2}+3^{1}a+3^{0}\right)\left(-9\right)a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$\left(\frac{1}{27}a^{3}+\frac{1}{9}a^{2}+3^{1}a+3^{0}\right)\left(-9\right)a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Calculate 3 to the power of 1 and get 3.

$$\left(\frac{1}{27}a^{3}+\frac{1}{9}a^{2}+3a+3^{0}\right)\left(-9\right)a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Calculate 3 to the power of 0 and get 1.

$$\left(\frac{1}{27}a^{3}+\frac{1}{9}a^{2}+3a+1\right)\left(-9\right)a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Use the distributive property to multiply \frac{1}{27}a^{3}+\frac{1}{9}a^{2}+3a+1 by -9.

$$\left(-\frac{1}{3}a^{3}-a^{2}-27a-9\right)a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Use the distributive property to multiply -\frac{1}{3}a^{3}-a^{2}-27a-9 by a^{2}.

$$-\frac{1}{3}a^{5}-a^{4}-27a^{3}-9a^{2}+a^{2}\left(a^{2}+27a+9\right)+1$$

Use the distributive property to multiply a^{2} by a^{2}+27a+9.

$$-\frac{1}{3}a^{5}-a^{4}-27a^{3}-9a^{2}+a^{4}+27a^{3}+9a^{2}+1$$

Combine -a^{4} and a^{4} to get 0.

$$-\frac{1}{3}a^{5}-27a^{3}-9a^{2}+27a^{3}+9a^{2}+1$$

Combine -27a^{3} and 27a^{3} to get 0.

$$-\frac{1}{3}a^{5}-9a^{2}+9a^{2}+1$$

Combine -9a^{2} and 9a^{2} to get 0.

$$-\frac{1}{3}a^{5}+1$$

Factor

$$\frac{3-a^{5}}{3}$$

Solution Steps

Factor out \frac{1}{3}. Polynomial -a^{5}+3 is not factored since it does not have any rational roots.

$$\frac{-\left(a^{3}+3a^{2}+81a+27\right)a^{2}+3a^{2}\left(a^{2}+27a+9\right)+3}{3}$$