Solve (b-2)^3-b*(3-b)^2

$$( b - 2 ) ^ { 3 } - b \cdot ( 3 - b ) ^ { 2 }$$

Evaluate

$$3b-8$$

Solution Steps

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

$$b^{3}-6b^{2}+12b-8-b\left(3-b\right)^{2}$$

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

$$b^{3}-6b^{2}+12b-8-b\left(9-6b+b^{2}\right)$$

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

$$b^{3}-6b^{2}+12b-8-\left(9b-6b^{2}+b^{3}\right)$$

To find the opposite of 9b-6b^{2}+b^{3}, find the opposite of each term.

$$b^{3}-6b^{2}+12b-8-9b+6b^{2}-b^{3}$$

Combine 12b and -9b to get 3b.

$$b^{3}-6b^{2}+3b-8+6b^{2}-b^{3}$$

Combine -6b^{2} and 6b^{2} to get 0.

$$b^{3}+3b-8-b^{3}$$

Combine b^{3} and -b^{3} to get 0.

$$3b-8$$

Expand

$$3b-8$$

Solution Steps

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

$$b^{3}-6b^{2}+12b-8-b\left(3-b\right)^{2}$$

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

$$b^{3}-6b^{2}+12b-8-b\left(9-6b+b^{2}\right)$$

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

$$b^{3}-6b^{2}+12b-8-\left(9b-6b^{2}+b^{3}\right)$$

To find the opposite of 9b-6b^{2}+b^{3}, find the opposite of each term.

$$b^{3}-6b^{2}+12b-8-9b+6b^{2}-b^{3}$$

Combine 12b and -9b to get 3b.

$$b^{3}-6b^{2}+3b-8+6b^{2}-b^{3}$$

Combine -6b^{2} and 6b^{2} to get 0.

$$b^{3}+3b-8-b^{3}$$

Combine b^{3} and -b^{3} to get 0.

$$3b-8$$