Solve (a-2)^3+(a-1)^3

$$( a - 2 ) ^ { 3 } + ( a - 1 ) ^ { 3 }$$

Evaluate

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

Solution Steps

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

$$a^{3}-6a^{2}+12a-8+\left(a-1\right)^{3}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a-1\right)^{3}.

$$a^{3}-6a^{2}+12a-8+a^{3}-3a^{2}+3a-1$$

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

$$2a^{3}-6a^{2}+12a-8-3a^{2}+3a-1$$

Combine -6a^{2} and -3a^{2} to get -9a^{2}.

$$2a^{3}-9a^{2}+12a-8+3a-1$$

Combine 12a and 3a to get 15a.

$$2a^{3}-9a^{2}+15a-8-1$$

Subtract 1 from -8 to get -9.

$$2a^{3}-9a^{2}+15a-9$$

Expand

$$2a^{3}-9a^{2}+15a-9$$

Solution Steps

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

$$a^{3}-6a^{2}+12a-8+\left(a-1\right)^{3}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(a-1\right)^{3}.

$$a^{3}-6a^{2}+12a-8+a^{3}-3a^{2}+3a-1$$

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

$$2a^{3}-6a^{2}+12a-8-3a^{2}+3a-1$$

Combine -6a^{2} and -3a^{2} to get -9a^{2}.

$$2a^{3}-9a^{2}+12a-8+3a-1$$

Combine 12a and 3a to get 15a.

$$2a^{3}-9a^{2}+15a-8-1$$

Subtract 1 from -8 to get -9.

$$2a^{3}-9a^{2}+15a-9$$