Solve a^2/3-3^2/2+a^2/6

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

Evaluate

$$\frac{a^{2}-9}{2}$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\frac{a^{2}}{3}-\frac{9}{2}+\frac{a^{2}}{6}$$

Combine \frac{a^{2}}{3} and \frac{a^{2}}{6} to get \frac{1}{2}a^{2}.

$$\frac{1}{2}a^{2}-\frac{9}{2}$$

Factor

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

Solution Steps

Factor out \frac{1}{6}.

$$\frac{2a^{2}-27+a^{2}}{6}$$

Consider 2a^{2}-27+a^{2}. Multiply and combine like terms.

$$3a^{2}-27$$

Consider 3a^{2}-27. Factor out 3.

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

Consider a^{2}-9. Rewrite a^{2}-9 as a^{2}-3^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-3\right)\left(a+3\right)$$

Rewrite the complete factored expression.

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