Solve (a+1/3x)^2-(a-1/2x)(a+1/2x)+11/36y^2=

$$( a + \frac { 1 } { 3 } x ) ^ { 2 } - ( a - \frac { 1 } { 2 } x ) ( a + \frac { 1 } { 2 } x ) + \frac { 11 } { 36 } y ^ { 2 } =$$

Evaluate

$$\frac{2ax}{3}+\frac{11y^{2}}{36}+\frac{13x^{2}}{36}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+\frac{1}{3}x\right)^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a-\frac{1}{2}x\right)\left(a+\frac{1}{2}x\right)+\frac{11}{36}y^{2}$$

Consider \left(a-\frac{1}{2}x\right)\left(a+\frac{1}{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\left(\frac{1}{2}x\right)^{2}\right)+\frac{11}{36}y^{2}$$

Expand \left(\frac{1}{2}x\right)^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\left(\frac{1}{2}\right)^{2}x^{2}\right)+\frac{11}{36}y^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\frac{1}{4}x^{2}\right)+\frac{11}{36}y^{2}$$

To find the opposite of a^{2}-\frac{1}{4}x^{2}, find the opposite of each term.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-a^{2}+\frac{1}{4}x^{2}+\frac{11}{36}y^{2}$$

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

$$\frac{2}{3}ax+\frac{1}{9}x^{2}+\frac{1}{4}x^{2}+\frac{11}{36}y^{2}$$

Combine \frac{1}{9}x^{2} and \frac{1}{4}x^{2} to get \frac{13}{36}x^{2}.

$$\frac{2}{3}ax+\frac{13}{36}x^{2}+\frac{11}{36}y^{2}$$

Expand

$$\frac{2ax}{3}+\frac{11y^{2}}{36}+\frac{13x^{2}}{36}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+\frac{1}{3}x\right)^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a-\frac{1}{2}x\right)\left(a+\frac{1}{2}x\right)+\frac{11}{36}y^{2}$$

Consider \left(a-\frac{1}{2}x\right)\left(a+\frac{1}{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\left(\frac{1}{2}x\right)^{2}\right)+\frac{11}{36}y^{2}$$

Expand \left(\frac{1}{2}x\right)^{2}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\left(\frac{1}{2}\right)^{2}x^{2}\right)+\frac{11}{36}y^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-\left(a^{2}-\frac{1}{4}x^{2}\right)+\frac{11}{36}y^{2}$$

To find the opposite of a^{2}-\frac{1}{4}x^{2}, find the opposite of each term.

$$a^{2}+\frac{2}{3}ax+\frac{1}{9}x^{2}-a^{2}+\frac{1}{4}x^{2}+\frac{11}{36}y^{2}$$

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

$$\frac{2}{3}ax+\frac{1}{9}x^{2}+\frac{1}{4}x^{2}+\frac{11}{36}y^{2}$$

Combine \frac{1}{9}x^{2} and \frac{1}{4}x^{2} to get \frac{13}{36}x^{2}.

$$\frac{2}{3}ax+\frac{13}{36}x^{2}+\frac{11}{36}y^{2}$$