Solve left(1-xright)^3-left(-x-1right)^3-6(x^2-2x+2)=

$${ \left(1-x \right) }^{ 3 } - { \left(-x-1 \right) }^{ 3 } -6( { x }^{ 2 } -2x+2)=$$

Evaluate

$$12x-10$$

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}.

$$1-3x+3x^{2}-x^{3}-\left(-x-1\right)^{3}-6\left(x^{2}-2x+2\right)$$

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

$$1-3x+3x^{2}-x^{3}-\left(\left(-x\right)^{3}-3\left(-x\right)^{2}+3\left(-x\right)-1\right)-6\left(x^{2}-2x+2\right)$$

Calculate -x to the power of 2 and get x^{2}.

$$1-3x+3x^{2}-x^{3}-\left(\left(-x\right)^{3}-3x^{2}+3\left(-x\right)-1\right)-6\left(x^{2}-2x+2\right)$$

To find the opposite of \left(-x\right)^{3}-3x^{2}+3\left(-x\right)-1, find the opposite of each term.

$$1-3x+3x^{2}-x^{3}-\left(-x\right)^{3}+3x^{2}-3\left(-x\right)+1-6\left(x^{2}-2x+2\right)$$

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

$$1-3x+6x^{2}-x^{3}-\left(-x\right)^{3}-3\left(-x\right)+1-6\left(x^{2}-2x+2\right)$$

Multiply -3 and -1 to get 3.

$$1-3x+6x^{2}-x^{3}-\left(-x\right)^{3}+3x+1-6\left(x^{2}-2x+2\right)$$

Combine -3x and 3x to get 0.

$$1+6x^{2}-x^{3}-\left(-x\right)^{3}+1-6\left(x^{2}-2x+2\right)$$

Add 1 and 1 to get 2.

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

Use the distributive property to multiply -6 by x^{2}-2x+2.

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

Combine 6x^{2} and -6x^{2} to get 0.

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

Subtract 12 from 2 to get -10.

$$-10-x^{3}-\left(-x\right)^{3}+12x$$

Expand \left(-x\right)^{3}.

$$-10-x^{3}-\left(-1\right)^{3}x^{3}+12x$$

Calculate -1 to the power of 3 and get -1.

$$-10-x^{3}-\left(-x^{3}\right)+12x$$

Multiply -1 and -1 to get 1.

$$-10-x^{3}+x^{3}+12x$$

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

$$-10+12x$$

Expand

$$12x-10$$

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}.

$$1-3x+3x^{2}-x^{3}-\left(-x-1\right)^{3}-6\left(x^{2}-2x+2\right)$$

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

$$1-3x+3x^{2}-x^{3}-\left(\left(-x\right)^{3}-3\left(-x\right)^{2}+3\left(-x\right)-1\right)-6\left(x^{2}-2x+2\right)$$

Calculate -x to the power of 2 and get x^{2}.

$$1-3x+3x^{2}-x^{3}-\left(\left(-x\right)^{3}-3x^{2}+3\left(-x\right)-1\right)-6\left(x^{2}-2x+2\right)$$

To find the opposite of \left(-x\right)^{3}-3x^{2}+3\left(-x\right)-1, find the opposite of each term.

$$1-3x+3x^{2}-x^{3}-\left(-x\right)^{3}+3x^{2}-3\left(-x\right)+1-6\left(x^{2}-2x+2\right)$$

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

$$1-3x+6x^{2}-x^{3}-\left(-x\right)^{3}-3\left(-x\right)+1-6\left(x^{2}-2x+2\right)$$

Multiply -3 and -1 to get 3.

$$1-3x+6x^{2}-x^{3}-\left(-x\right)^{3}+3x+1-6\left(x^{2}-2x+2\right)$$

Combine -3x and 3x to get 0.

$$1+6x^{2}-x^{3}-\left(-x\right)^{3}+1-6\left(x^{2}-2x+2\right)$$

Add 1 and 1 to get 2.

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

Use the distributive property to multiply -6 by x^{2}-2x+2.

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

Combine 6x^{2} and -6x^{2} to get 0.

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

Subtract 12 from 2 to get -10.

$$-10-x^{3}-\left(-x\right)^{3}+12x$$

Expand \left(-x\right)^{3}.

$$-10-x^{3}-\left(-1\right)^{3}x^{3}+12x$$

Calculate -1 to the power of 3 and get -1.

$$-10-x^{3}-\left(-x^{3}\right)+12x$$

Multiply -1 and -1 to get 1.

$$-10-x^{3}+x^{3}+12x$$

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

$$-10+12x$$