Solve (y(y+1)(2y+1)1/6-1)

$$(y(y+1)(2y+1) \frac{ 1 }{ 6 } -1)$$

Evaluate

$$\frac{y\left(y+1\right)\left(2y+1\right)-6}{6}$$

Solution Steps

Use the distributive property to multiply y by y+1.

$$\left(y^{2}+y\right)\left(2y+1\right)\times \frac{1}{6}-1$$

Apply the distributive property by multiplying each term of y^{2}+y by each term of 2y+1.

$$\left(2y^{3}+y^{2}+2y^{2}+y\right)\times \frac{1}{6}-1$$

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

$$\left(2y^{3}+3y^{2}+y\right)\times \frac{1}{6}-1$$

Use the distributive property to multiply 2y^{3}+3y^{2}+y by \frac{1}{6}.

$$2y^{3}\times \frac{1}{6}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Multiply 2 and \frac{1}{6} to get \frac{2}{6}.

$$\frac{2}{6}y^{3}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$\frac{1}{3}y^{3}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Multiply 3 and \frac{1}{6} to get \frac{3}{6}.

$$\frac{1}{3}y^{3}+\frac{3}{6}y^{2}+y\times \frac{1}{6}-1$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{1}{3}y^{3}+\frac{1}{2}y^{2}+y\times \frac{1}{6}-1$$

Expand

$$\frac{y^{3}}{3}+\frac{y^{2}}{2}+\frac{y}{6}-1$$

Solution Steps

Use the distributive property to multiply y by y+1.

$$\left(y^{2}+y\right)\left(2y+1\right)\times \frac{1}{6}-1$$

Apply the distributive property by multiplying each term of y^{2}+y by each term of 2y+1.

$$\left(2y^{3}+y^{2}+2y^{2}+y\right)\times \frac{1}{6}-1$$

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

$$\left(2y^{3}+3y^{2}+y\right)\times \frac{1}{6}-1$$

Use the distributive property to multiply 2y^{3}+3y^{2}+y by \frac{1}{6}.

$$2y^{3}\times \frac{1}{6}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Multiply 2 and \frac{1}{6} to get \frac{2}{6}.

$$\frac{2}{6}y^{3}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$\frac{1}{3}y^{3}+3y^{2}\times \frac{1}{6}+y\times \frac{1}{6}-1$$

Multiply 3 and \frac{1}{6} to get \frac{3}{6}.

$$\frac{1}{3}y^{3}+\frac{3}{6}y^{2}+y\times \frac{1}{6}-1$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{1}{3}y^{3}+\frac{1}{2}y^{2}+y\times \frac{1}{6}-1$$