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

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

Evaluate

$$8$$

Solution Steps

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

$$-8-12x-6x^{2}-x^{3}+3\left(-2-x\right)^{2}+2+x^{3}+3x^{2}+2$$

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

$$-8-12x-6x^{2}-x^{3}+3\left(4+4x+x^{2}\right)+2+x^{3}+3x^{2}+2$$

Use the distributive property to multiply 3 by 4+4x+x^{2}.

$$-8-12x-6x^{2}-x^{3}+12+12x+3x^{2}+2+x^{3}+3x^{2}+2$$

Add -8 and 12 to get 4.

$$4-12x-6x^{2}-x^{3}+12x+3x^{2}+2+x^{3}+3x^{2}+2$$

Combine -12x and 12x to get 0.

$$4-6x^{2}-x^{3}+3x^{2}+2+x^{3}+3x^{2}+2$$

Combine -6x^{2} and 3x^{2} to get -3x^{2}.

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

Add 4 and 2 to get 6.

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

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

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

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

$$6+2$$

Add 6 and 2 to get 8.

$$8$$

Factor

$$2^{3}$$