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

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

Evaluate

$$4-4x^{3}$$

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}.

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

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

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

Use the distributive property to multiply 3 by 1-x.

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

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

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

Add 1 and 3 to get 4.

$$4-3x+3x^{2}-x^{3}+3x-3x^{2}-3x^{3}$$

Combine -3x and 3x to get 0.

$$4+3x^{2}-x^{3}-3x^{2}-3x^{3}$$

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

$$4-x^{3}-3x^{3}$$

Combine -x^{3} and -3x^{3} to get -4x^{3}.

$$4-4x^{3}$$

Expand

$$4-4x^{3}$$

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}.

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

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

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

Use the distributive property to multiply 3 by 1-x.

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

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

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

Add 1 and 3 to get 4.

$$4-3x+3x^{2}-x^{3}+3x-3x^{2}-3x^{3}$$

Combine -3x and 3x to get 0.

$$4+3x^{2}-x^{3}-3x^{2}-3x^{3}$$

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

$$4-x^{3}-3x^{3}$$

Combine -x^{3} and -3x^{3} to get -4x^{3}.

$$4-4x^{3}$$