Solve (3a+b-2c)(-3a+b-2c)

$$( 3 a + b - 2 c ) ( - 3 a + b - 2 c )$$

Evaluate

$$\left(b-2c\right)^{2}-9a^{2}$$

Solution Steps

Apply the distributive property by multiplying each term of 3a+b-2c by each term of -3a+b-2c.

$$-9a^{2}+3ab-6ac-3ba+b^{2}-2bc+6ca-2cb+4c^{2}$$

Combine 3ab and -3ba to get 0.

$$-9a^{2}-6ac+b^{2}-2bc+6ca-2cb+4c^{2}$$

Combine -6ac and 6ca to get 0.

$$-9a^{2}+b^{2}-2bc-2cb+4c^{2}$$

Combine -2bc and -2cb to get -4bc.

$$-9a^{2}+b^{2}-4bc+4c^{2}$$

Expand

$$4c^{2}-4bc+b^{2}-9a^{2}$$

Solution Steps

Apply the distributive property by multiplying each term of 3a+b-2c by each term of -3a+b-2c.

$$-9a^{2}+3ab-6ac-3ba+b^{2}-2bc+6ca-2cb+4c^{2}$$

Combine 3ab and -3ba to get 0.

$$-9a^{2}-6ac+b^{2}-2bc+6ca-2cb+4c^{2}$$

Combine -6ac and 6ca to get 0.

$$-9a^{2}+b^{2}-2bc-2cb+4c^{2}$$

Combine -2bc and -2cb to get -4bc.

$$-9a^{2}+b^{2}-4bc+4c^{2}$$