Solve (0,2x-1)(0,2x+1)-(1/5x-1)^2+(x+1)^3-(x-1)^3-2/5x

$$( 0,2 x - 1 ) ( 0,2 x + 1 ) - ( \frac { 1 } { 5 } x - 1 ) ^ { 2 } + ( x + 1 ) ^ { 3 } - ( x - 1 ) ^ { 3 } - \frac { 2 } { 5 } x$$

Evaluate

$$6x^{2}$$

Solution Steps

Consider \left(0,2x-1\right)\left(0,2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(0,2x\right)^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Expand \left(0,2x\right)^{2}.

$$0,2^{2}x^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Calculate 0,2 to the power of 2 and get 0,04.

$$0,04x^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}x-1\right)^{2}.

$$0,04x^{2}-1-\left(\frac{1}{25}x^{2}-\frac{2}{5}x+1\right)+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

To find the opposite of \frac{1}{25}x^{2}-\frac{2}{5}x+1, find the opposite of each term.

$$0,04x^{2}-1-\frac{1}{25}x^{2}+\frac{2}{5}x-1+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Combine 0,04x^{2} and -\frac{1}{25}x^{2} to get 0.

$$-1+\frac{2}{5}x-1+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Subtract 1 from -1 to get -2.

$$-2+\frac{2}{5}x+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+1\right)^{3}.

$$-2+\frac{2}{5}x+x^{3}+3x^{2}+3x+1-\left(x-1\right)^{3}-\frac{2}{5}x$$

Combine \frac{2}{5}x and 3x to get \frac{17}{5}x.

$$-2+\frac{17}{5}x+x^{3}+3x^{2}+1-\left(x-1\right)^{3}-\frac{2}{5}x$$

Add -2 and 1 to get -1.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-\left(x^{3}-3x^{2}+3x-1\right)-\frac{2}{5}x$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-x^{3}+3x^{2}-3x+1-\frac{2}{5}x$$

Combine x^{3} and -x^{3} to get 0.

$$-1+\frac{17}{5}x+3x^{2}+3x^{2}-3x+1-\frac{2}{5}x$$

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

$$-1+\frac{17}{5}x+6x^{2}-3x+1-\frac{2}{5}x$$

Combine \frac{17}{5}x and -3x to get \frac{2}{5}x.

$$-1+\frac{2}{5}x+6x^{2}+1-\frac{2}{5}x$$

Add -1 and 1 to get 0.

$$\frac{2}{5}x+6x^{2}-\frac{2}{5}x$$

Combine \frac{2}{5}x and -\frac{2}{5}x to get 0.

$$6x^{2}$$

Expand

$$6x^{2}$$

Solution Steps

Consider \left(0,2x-1\right)\left(0,2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(0,2x\right)^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Expand \left(0,2x\right)^{2}.

$$0,2^{2}x^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Calculate 0,2 to the power of 2 and get 0,04.

$$0,04x^{2}-1-\left(\frac{1}{5}x-1\right)^{2}+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}x-1\right)^{2}.

$$0,04x^{2}-1-\left(\frac{1}{25}x^{2}-\frac{2}{5}x+1\right)+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

To find the opposite of \frac{1}{25}x^{2}-\frac{2}{5}x+1, find the opposite of each term.

$$0,04x^{2}-1-\frac{1}{25}x^{2}+\frac{2}{5}x-1+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Combine 0,04x^{2} and -\frac{1}{25}x^{2} to get 0.

$$-1+\frac{2}{5}x-1+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Subtract 1 from -1 to get -2.

$$-2+\frac{2}{5}x+\left(x+1\right)^{3}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+1\right)^{3}.

$$-2+\frac{2}{5}x+x^{3}+3x^{2}+3x+1-\left(x-1\right)^{3}-\frac{2}{5}x$$

Combine \frac{2}{5}x and 3x to get \frac{17}{5}x.

$$-2+\frac{17}{5}x+x^{3}+3x^{2}+1-\left(x-1\right)^{3}-\frac{2}{5}x$$

Add -2 and 1 to get -1.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-\left(x-1\right)^{3}-\frac{2}{5}x$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-\left(x^{3}-3x^{2}+3x-1\right)-\frac{2}{5}x$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$-1+\frac{17}{5}x+x^{3}+3x^{2}-x^{3}+3x^{2}-3x+1-\frac{2}{5}x$$

Combine x^{3} and -x^{3} to get 0.

$$-1+\frac{17}{5}x+3x^{2}+3x^{2}-3x+1-\frac{2}{5}x$$

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

$$-1+\frac{17}{5}x+6x^{2}-3x+1-\frac{2}{5}x$$

Combine \frac{17}{5}x and -3x to get \frac{2}{5}x.

$$-1+\frac{2}{5}x+6x^{2}+1-\frac{2}{5}x$$

Add -1 and 1 to get 0.

$$\frac{2}{5}x+6x^{2}-\frac{2}{5}x$$

Combine \frac{2}{5}x and -\frac{2}{5}x to get 0.

$$6x^{2}$$