Solve (2a-x)(2a+x)+(a-2x)^2-(x-2a)^2

$$( 2 a - x ) ( 2 a + x ) + ( a - 2 x ) ^ { 2 } - ( x - 2 a ) ^ { 2 }$$

Evaluate

$$2x^{2}+a^{2}$$

Solution Steps

Consider \left(2a-x\right)\left(2a+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}.

$$\left(2a\right)^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

Expand \left(2a\right)^{2}.

$$2^{2}a^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4a^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

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

$$4a^{2}-x^{2}+a^{2}-4ax+4x^{2}-\left(x-2a\right)^{2}$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}-x^{2}-4ax+4x^{2}-\left(x-2a\right)^{2}$$

Combine -x^{2} and 4x^{2} to get 3x^{2}.

$$5a^{2}+3x^{2}-4ax-\left(x-2a\right)^{2}$$

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

$$5a^{2}+3x^{2}-4ax-\left(x^{2}-4xa+4a^{2}\right)$$

To find the opposite of x^{2}-4xa+4a^{2}, find the opposite of each term.

$$5a^{2}+3x^{2}-4ax-x^{2}+4xa-4a^{2}$$

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

$$5a^{2}+2x^{2}-4ax+4xa-4a^{2}$$

Combine -4ax and 4xa to get 0.

$$5a^{2}+2x^{2}-4a^{2}$$

Combine 5a^{2} and -4a^{2} to get a^{2}.

$$a^{2}+2x^{2}$$

Expand

$$2x^{2}+a^{2}$$

Solution Steps

Consider \left(2a-x\right)\left(2a+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}.

$$\left(2a\right)^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

Expand \left(2a\right)^{2}.

$$2^{2}a^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4a^{2}-x^{2}+\left(a-2x\right)^{2}-\left(x-2a\right)^{2}$$

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

$$4a^{2}-x^{2}+a^{2}-4ax+4x^{2}-\left(x-2a\right)^{2}$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}-x^{2}-4ax+4x^{2}-\left(x-2a\right)^{2}$$

Combine -x^{2} and 4x^{2} to get 3x^{2}.

$$5a^{2}+3x^{2}-4ax-\left(x-2a\right)^{2}$$

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

$$5a^{2}+3x^{2}-4ax-\left(x^{2}-4xa+4a^{2}\right)$$

To find the opposite of x^{2}-4xa+4a^{2}, find the opposite of each term.

$$5a^{2}+3x^{2}-4ax-x^{2}+4xa-4a^{2}$$

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

$$5a^{2}+2x^{2}-4ax+4xa-4a^{2}$$

Combine -4ax and 4xa to get 0.

$$5a^{2}+2x^{2}-4a^{2}$$

Combine 5a^{2} and -4a^{2} to get a^{2}.

$$a^{2}+2x^{2}$$