Solve (a-3b)^2-9(a-3b)^2=

$$( a - 3 b ) ^ { 2 } - 9 ( a - 3 b ) ^ { 2 } =$$

Evaluate

$$-8\left(a-3b\right)^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3b\right)^{2}.

$$a^{2}-6ab+9b^{2}-9\left(a-3b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3b\right)^{2}.

$$a^{2}-6ab+9b^{2}-9\left(a^{2}-6ab+9b^{2}\right)$$

Use the distributive property to multiply -9 by a^{2}-6ab+9b^{2}.

$$a^{2}-6ab+9b^{2}-9a^{2}+54ab-81b^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-6ab+9b^{2}+54ab-81b^{2}$$

Combine -6ab and 54ab to get 48ab.

$$-8a^{2}+48ab+9b^{2}-81b^{2}$$

Combine 9b^{2} and -81b^{2} to get -72b^{2}.

$$-8a^{2}+48ab-72b^{2}$$

Expand

$$-8a^{2}+48ab-72b^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3b\right)^{2}.

$$a^{2}-6ab+9b^{2}-9\left(a-3b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3b\right)^{2}.

$$a^{2}-6ab+9b^{2}-9\left(a^{2}-6ab+9b^{2}\right)$$

Use the distributive property to multiply -9 by a^{2}-6ab+9b^{2}.

$$a^{2}-6ab+9b^{2}-9a^{2}+54ab-81b^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-6ab+9b^{2}+54ab-81b^{2}$$

Combine -6ab and 54ab to get 48ab.

$$-8a^{2}+48ab+9b^{2}-81b^{2}$$

Combine 9b^{2} and -81b^{2} to get -72b^{2}.

$$-8a^{2}+48ab-72b^{2}$$