Solve (x-2)(x+2)-(x-1)^{3}-4x^{2}]^{2}-(x^3-3)^2-6x(x^3+2x+3)+3x^2

$$( x - 2 ) ( x + 2 ) - ( x - 1 ) ^ { 3 } - 4 x ^ { 2 } ] ^ { 2 } - ( x ^ { 3 } - 3 ) ^ { 2 } - 6 x ( x ^ { 3 } + 2 x + 3 ) + 3 x ^ { 2 }$$

Evaluate

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

Solution Steps

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$x^{2}-4-\left(x-1\right)^{3}-4x^{2}$$

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

$$x^{2}-4-\left(x^{3}-3x^{2}+3x-1\right)-4x^{2}$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$x^{2}-4-x^{3}+3x^{2}-3x+1-4x^{2}$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-4-x^{3}-3x+1-4x^{2}$$

Add -4 and 1 to get -3.

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

Combine 4x^{2} and -4x^{2} to get 0.

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

Expand

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

Solution Steps

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$x^{2}-4-\left(x-1\right)^{3}-4x^{2}$$

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

$$x^{2}-4-\left(x^{3}-3x^{2}+3x-1\right)-4x^{2}$$

To find the opposite of x^{3}-3x^{2}+3x-1, find the opposite of each term.

$$x^{2}-4-x^{3}+3x^{2}-3x+1-4x^{2}$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-4-x^{3}-3x+1-4x^{2}$$

Add -4 and 1 to get -3.

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

Combine 4x^{2} and -4x^{2} to get 0.

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