Solve (4a-3)^2+4(3-4a)^3

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

Evaluate

$$\left(13-16a\right)\left(3-4a\right)^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-3\right)^{2}.

$$16a^{2}-24a+9+4\left(3-4a\right)^{3}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(3-4a\right)^{3}.

$$16a^{2}-24a+9+4\left(27-108a+144a^{2}-64a^{3}\right)$$

Use the distributive property to multiply 4 by 27-108a+144a^{2}-64a^{3}.

$$16a^{2}-24a+9+108-432a+576a^{2}-256a^{3}$$

Add 9 and 108 to get 117.

$$16a^{2}-24a+117-432a+576a^{2}-256a^{3}$$

Combine -24a and -432a to get -456a.

$$16a^{2}-456a+117+576a^{2}-256a^{3}$$

Combine 16a^{2} and 576a^{2} to get 592a^{2}.

$$592a^{2}-456a+117-256a^{3}$$

Expand

$$117-456a+592a^{2}-256a^{3}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-3\right)^{2}.

$$16a^{2}-24a+9+4\left(3-4a\right)^{3}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(3-4a\right)^{3}.

$$16a^{2}-24a+9+4\left(27-108a+144a^{2}-64a^{3}\right)$$

Use the distributive property to multiply 4 by 27-108a+144a^{2}-64a^{3}.

$$16a^{2}-24a+9+108-432a+576a^{2}-256a^{3}$$

Add 9 and 108 to get 117.

$$16a^{2}-24a+117-432a+576a^{2}-256a^{3}$$

Combine -24a and -432a to get -456a.

$$16a^{2}-456a+117+576a^{2}-256a^{3}$$

Combine 16a^{2} and 576a^{2} to get 592a^{2}.

$$592a^{2}-456a+117-256a^{3}$$