Solve [(x-1)^{3}-(x+1)^3]^2-4(3x^2+1)(3x^2-1)-8=

$$[ ( x - 1 ) ^ { 3 } - ( x + 1 ) ^ { 3 } ] ^ { 2 } - 4 ( 3 x ^ { 2 } + 1 ) ( 3 x ^ { 2 } - 1 ) - 8 =$$

Evaluate

$$24x^{2}$$

Solution Steps

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

$$\left(x^{3}-3x^{2}+3x-1-\left(x+1\right)^{3}\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(x^{3}-3x^{2}+3x-1-\left(x^{3}+3x^{2}+3x+1\right)\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(x^{3}-3x^{2}+3x-1-x^{3}-3x^{2}-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(-3x^{2}+3x-1-3x^{2}-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(-6x^{2}+3x-1-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Combine 3x and -3x to get 0.

$$\left(-6x^{2}-1-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Subtract 1 from -1 to get -2.

$$\left(-6x^{2}-2\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6x^{2}-2\right)^{2}.

$$36\left(x^{2}\right)^{2}+24x^{2}+4-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36x^{4}+24x^{2}+4-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Use the distributive property to multiply -4 by 3x^{2}+1.

$$36x^{4}+24x^{2}+4+\left(-12x^{2}-4\right)\left(3x^{2}-1\right)-8$$

Use the distributive property to multiply -12x^{2}-4 by 3x^{2}-1 and combine like terms.

$$36x^{4}+24x^{2}+4-36x^{4}+4-8$$

Combine 36x^{4} and -36x^{4} to get 0.

$$24x^{2}+4+4-8$$

Add 4 and 4 to get 8.

$$24x^{2}+8-8$$

Subtract 8 from 8 to get 0.

$$24x^{2}$$

Expand

$$24x^{2}$$

Solution Steps

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

$$\left(x^{3}-3x^{2}+3x-1-\left(x+1\right)^{3}\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(x^{3}-3x^{2}+3x-1-\left(x^{3}+3x^{2}+3x+1\right)\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(x^{3}-3x^{2}+3x-1-x^{3}-3x^{2}-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(-3x^{2}+3x-1-3x^{2}-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

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

$$\left(-6x^{2}+3x-1-3x-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Combine 3x and -3x to get 0.

$$\left(-6x^{2}-1-1\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Subtract 1 from -1 to get -2.

$$\left(-6x^{2}-2\right)^{2}-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6x^{2}-2\right)^{2}.

$$36\left(x^{2}\right)^{2}+24x^{2}+4-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36x^{4}+24x^{2}+4-4\left(3x^{2}+1\right)\left(3x^{2}-1\right)-8$$

Use the distributive property to multiply -4 by 3x^{2}+1.

$$36x^{4}+24x^{2}+4+\left(-12x^{2}-4\right)\left(3x^{2}-1\right)-8$$

Use the distributive property to multiply -12x^{2}-4 by 3x^{2}-1 and combine like terms.

$$36x^{4}+24x^{2}+4-36x^{4}+4-8$$

Combine 36x^{4} and -36x^{4} to get 0.

$$24x^{2}+4+4-8$$

Add 4 and 4 to get 8.

$$24x^{2}+8-8$$

Subtract 8 from 8 to get 0.

$$24x^{2}$$