Solve 1/3(8-(4-y)^3)

$$\frac { 1 } { 3 } ( 8 - ( 4 - y ) ^ { 3 } )$$

Evaluate

$$\frac{\left(y-2\right)\left(y^{2}-10y+28\right)}{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4-y\right)^{3}.

$$\frac{1}{3}\left(8-\left(64-48y+12y^{2}-y^{3}\right)\right)$$

To find the opposite of 64-48y+12y^{2}-y^{3}, find the opposite of each term.

$$\frac{1}{3}\left(8-64+48y-12y^{2}+y^{3}\right)$$

Subtract 64 from 8 to get -56.

$$\frac{1}{3}\left(-56+48y-12y^{2}+y^{3}\right)$$

Use the distributive property to multiply \frac{1}{3} by -56+48y-12y^{2}+y^{3}.

$$-\frac{56}{3}+16y-4y^{2}+\frac{1}{3}y^{3}$$

Expand

$$\frac{y^{3}}{3}-4y^{2}+16y-\frac{56}{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4-y\right)^{3}.

$$\frac{1}{3}\left(8-\left(64-48y+12y^{2}-y^{3}\right)\right)$$

To find the opposite of 64-48y+12y^{2}-y^{3}, find the opposite of each term.

$$\frac{1}{3}\left(8-64+48y-12y^{2}+y^{3}\right)$$

Subtract 64 from 8 to get -56.

$$\frac{1}{3}\left(-56+48y-12y^{2}+y^{3}\right)$$

Use the distributive property to multiply \frac{1}{3} by -56+48y-12y^{2}+y^{3}.

$$-\frac{56}{3}+16y-4y^{2}+\frac{1}{3}y^{3}$$