Solve (a+b)^2-[(a-b)^2+(a+b)(a-b)-4a(a-b)]-(3a^2+b^2)

$$( a + b ) ^ { 2 } - [ ( a - b ) ^ { 2 } + ( a + b ) ( a - b ) - 4 a ( a - b ) ] - ( 3 a ^ { 2 } + b ^ { 2 } )$$

Evaluate

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Solution Steps

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

$$a^{2}+2ab+b^{2}-\left(\left(a-b\right)^{2}+\left(a+b\right)\left(a-b\right)-4a\left(a-b\right)\right)-\left(3a^{2}+b^{2}\right)$$

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

$$a^{2}+2ab+b^{2}-\left(a^{2}-2ab+b^{2}+\left(a+b\right)\left(a-b\right)-4a\left(a-b\right)\right)-\left(3a^{2}+b^{2}\right)$$

Consider \left(a+b\right)\left(a-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}.

$$a^{2}+2ab+b^{2}-\left(a^{2}-2ab+b^{2}+a^{2}-b^{2}-4a\left(a-b\right)\right)-\left(3a^{2}+b^{2}\right)$$

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

$$a^{2}+2ab+b^{2}-\left(2a^{2}-2ab+b^{2}-b^{2}-4a\left(a-b\right)\right)-\left(3a^{2}+b^{2}\right)$$

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

$$a^{2}+2ab+b^{2}-\left(2a^{2}-2ab-4a\left(a-b\right)\right)-\left(3a^{2}+b^{2}\right)$$

To find the opposite of 3a^{2}+b^{2}, find the opposite of each term.

$$a^{2}+2ab+b^{2}-\left(2a^{2}-2ab-4a\left(a-b\right)\right)-3a^{2}-b^{2}$$

Use the distributive property to multiply -4a by a-b.

$$a^{2}+2ab+b^{2}-\left(2a^{2}-2ab-4a^{2}+4ab\right)-3a^{2}-b^{2}$$

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

$$a^{2}+2ab+b^{2}-\left(-2a^{2}-2ab+4ab\right)-3a^{2}-b^{2}$$

Combine -2ab and 4ab to get 2ab.

$$a^{2}+2ab+b^{2}-\left(-2a^{2}+2ab\right)-3a^{2}-b^{2}$$

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

$$a^{2}+2ab+b^{2}+2a^{2}-2ab-3a^{2}-b^{2}$$

Combine a^{2} and 2a^{2} to get 3a^{2}.

$$3a^{2}+2ab+b^{2}-2ab-3a^{2}-b^{2}$$

Combine 2ab and -2ab to get 0.

$$3a^{2}+b^{2}-3a^{2}-b^{2}$$

Combine 3a^{2} and -3a^{2} to get 0.

$$b^{2}-b^{2}$$

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

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Factor

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