Solve (1/2x+7/3y^2)(1/2x-7/3y^2)

$$( \frac { 1 } { 2 } x + \frac { 7 } { 3 } y ^ { 2 } ) ( \frac { 1 } { 2 } x - \frac { 7 } { 3 } y ^ { 2 } )$$

Evaluate

$$-\frac{49y^{4}}{9}+\frac{x^{2}}{4}$$

Solution Steps

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

$$\left(\frac{1}{2}x\right)^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

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

$$\left(\frac{1}{2}\right)^{2}x^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

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

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

Expand \left(\frac{7}{3}y^{2}\right)^{2}.

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}\right)^{2}\left(y^{2}\right)^{2}$$

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

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}\right)^{2}y^{4}$$

Calculate \frac{7}{3} to the power of 2 and get \frac{49}{9}.

$$\frac{1}{4}x^{2}-\frac{49}{9}y^{4}$$

Expand

$$-\frac{49y^{4}}{9}+\frac{x^{2}}{4}$$

Solution Steps

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

$$\left(\frac{1}{2}x\right)^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

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

$$\left(\frac{1}{2}\right)^{2}x^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

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

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}y^{2}\right)^{2}$$

Expand \left(\frac{7}{3}y^{2}\right)^{2}.

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}\right)^{2}\left(y^{2}\right)^{2}$$

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

$$\frac{1}{4}x^{2}-\left(\frac{7}{3}\right)^{2}y^{4}$$

Calculate \frac{7}{3} to the power of 2 and get \frac{49}{9}.

$$\frac{1}{4}x^{2}-\frac{49}{9}y^{4}$$