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

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

Evaluate

$$27-78n+81n^{2}-27n^{3}$$

Solution Steps

Calculate 1 to the power of 3 and get 1.

$$n+1+2n-1-\left(n+2n-3\right)^{3}$$

Combine n and 2n to get 3n.

$$3n+1-1-\left(n+2n-3\right)^{3}$$

Subtract 1 from 1 to get 0.

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

Combine n and 2n to get 3n.

$$3n-\left(3n-3\right)^{3}$$

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

$$3n-\left(27n^{3}-81n^{2}+81n-27\right)$$

To find the opposite of 27n^{3}-81n^{2}+81n-27, find the opposite of each term.

$$3n-27n^{3}+81n^{2}-81n+27$$

Combine 3n and -81n to get -78n.

$$-78n-27n^{3}+81n^{2}+27$$

Expand

$$27-78n+81n^{2}-27n^{3}$$

Solution Steps

Calculate 1 to the power of 3 and get 1.

$$n+1+2n-1-\left(n+2n-3\right)^{3}$$

Combine n and 2n to get 3n.

$$3n+1-1-\left(n+2n-3\right)^{3}$$

Subtract 1 from 1 to get 0.

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

Combine n and 2n to get 3n.

$$3n-\left(3n-3\right)^{3}$$

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

$$3n-\left(27n^{3}-81n^{2}+81n-27\right)$$

To find the opposite of 27n^{3}-81n^{2}+81n-27, find the opposite of each term.

$$3n-27n^{3}+81n^{2}-81n+27$$

Combine 3n and -81n to get -78n.

$$-78n-27n^{3}+81n^{2}+27$$