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

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

Evaluate

$$x^{3}-3x-1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+x\right)^{3}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

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

Use the distributive property to multiply -3 by 1+2x+x^{2}.

$$1+3x+3x^{2}+x^{3}-3-6x-3x^{2}+1$$

Subtract 3 from 1 to get -2.

$$-2+3x+3x^{2}+x^{3}-6x-3x^{2}+1$$

Combine 3x and -6x to get -3x.

$$-2-3x+3x^{2}+x^{3}-3x^{2}+1$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-2-3x+x^{3}+1$$

Add -2 and 1 to get -1.

$$-1-3x+x^{3}$$

Expand

$$x^{3}-3x-1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+x\right)^{3}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

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

Use the distributive property to multiply -3 by 1+2x+x^{2}.

$$1+3x+3x^{2}+x^{3}-3-6x-3x^{2}+1$$

Subtract 3 from 1 to get -2.

$$-2+3x+3x^{2}+x^{3}-6x-3x^{2}+1$$

Combine 3x and -6x to get -3x.

$$-2-3x+3x^{2}+x^{3}-3x^{2}+1$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-2-3x+x^{3}+1$$

Add -2 and 1 to get -1.

$$-1-3x+x^{3}$$