Solve 1/3k(k^2-1)+(k+7)^2-(k+1)

$$\frac { 1 } { 3 } k ( k ^ { 2 } - 1 ) + ( k + 7 ) ^ { 2 } - ( k + 1 )$$

Evaluate

$$\frac{k^{3}}{3}+k^{2}+\frac{38k}{3}+48$$

Solution Steps

Use the distributive property to multiply \frac{1}{3}k by k^{2}-1.

$$\frac{1}{3}k^{3}-\frac{1}{3}k+\left(k+7\right)^{2}-\left(k+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+7\right)^{2}.

$$\frac{1}{3}k^{3}-\frac{1}{3}k+k^{2}+14k+49-\left(k+1\right)$$

Combine -\frac{1}{3}k and 14k to get \frac{41}{3}k.

$$\frac{1}{3}k^{3}+\frac{41}{3}k+k^{2}+49-\left(k+1\right)$$

To find the opposite of k+1, find the opposite of each term.

$$\frac{1}{3}k^{3}+\frac{41}{3}k+k^{2}+49-k-1$$

Combine \frac{41}{3}k and -k to get \frac{38}{3}k.

$$\frac{1}{3}k^{3}+\frac{38}{3}k+k^{2}+49-1$$

Subtract 1 from 49 to get 48.

$$\frac{1}{3}k^{3}+\frac{38}{3}k+k^{2}+48$$

Expand

$$\frac{k^{3}}{3}+k^{2}+\frac{38k}{3}+48$$

Solution Steps

Use the distributive property to multiply \frac{1}{3}k by k^{2}-1.

$$\frac{1}{3}k^{3}-\frac{1}{3}k+\left(k+7\right)^{2}-\left(k+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+7\right)^{2}.

$$\frac{1}{3}k^{3}-\frac{1}{3}k+k^{2}+14k+49-\left(k+1\right)$$

Combine -\frac{1}{3}k and 14k to get \frac{41}{3}k.

$$\frac{1}{3}k^{3}+\frac{41}{3}k+k^{2}+49-\left(k+1\right)$$

To find the opposite of k+1, find the opposite of each term.

$$\frac{1}{3}k^{3}+\frac{41}{3}k+k^{2}+49-k-1$$

Combine \frac{41}{3}k and -k to get \frac{38}{3}k.

$$\frac{1}{3}k^{3}+\frac{38}{3}k+k^{2}+49-1$$

Subtract 1 from 49 to get 48.

$$\frac{1}{3}k^{3}+\frac{38}{3}k+k^{2}+48$$