Solve [(1-a)^2+(2-b)^2]+(2a-b)(2a+b)+(1+a)^2+4b)-(a-b)(a+b)

$$[ ( 1 - a ) ^ { 2 } + ( 2 - b ) ^ { 2 } ] + ( 2 a - b ) ( 2 a + b ) + ( 1 + a ) ^ { 2 } + 4 b ) - ( a - b ) ( a + b )$$

Evaluate

$$6\left(a^{2}+1\right)$$

Solution Steps

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

$$1-2a+a^{2}+\left(2-b\right)^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

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

$$1-2a+a^{2}+4-4b+b^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

Add 1 and 4 to get 5.

$$5-2a+a^{2}-4b+b^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

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

$$5-2a+a^{2}-4b+b^{2}+\left(2a\right)^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+a^{2}-4b+b^{2}+2^{2}a^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

Calculate 2 to the power of 2 and get 4.

$$5-2a+a^{2}-4b+b^{2}+4a^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+b^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+1+2a+a^{2}+4b$$

Add 5 and 1 to get 6.

$$6-2a+5a^{2}-4b+2a+a^{2}+4b$$

Combine -2a and 2a to get 0.

$$6+5a^{2}-4b+a^{2}+4b$$

Combine 5a^{2} and a^{2} to get 6a^{2}.

$$6+6a^{2}-4b+4b$$

Combine -4b and 4b to get 0.

$$6+6a^{2}$$

Expand

$$6a^{2}+6$$

Solution Steps

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

$$1-2a+a^{2}+\left(2-b\right)^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

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

$$1-2a+a^{2}+4-4b+b^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

Add 1 and 4 to get 5.

$$5-2a+a^{2}-4b+b^{2}+\left(2a-b\right)\left(2a+b\right)+\left(1+a\right)^{2}+4b$$

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

$$5-2a+a^{2}-4b+b^{2}+\left(2a\right)^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+a^{2}-4b+b^{2}+2^{2}a^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

Calculate 2 to the power of 2 and get 4.

$$5-2a+a^{2}-4b+b^{2}+4a^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+b^{2}-b^{2}+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+\left(1+a\right)^{2}+4b$$

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

$$5-2a+5a^{2}-4b+1+2a+a^{2}+4b$$

Add 5 and 1 to get 6.

$$6-2a+5a^{2}-4b+2a+a^{2}+4b$$

Combine -2a and 2a to get 0.

$$6+5a^{2}-4b+a^{2}+4b$$

Combine 5a^{2} and a^{2} to get 6a^{2}.

$$6+6a^{2}-4b+4b$$

Combine -4b and 4b to get 0.

$$6+6a^{2}$$