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

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

Evaluate

$$-2x^{3}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

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

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

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

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

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

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

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

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

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

To find the opposite of x^{6}+2x^{3}+1, find the opposite of each term.

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

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

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

Subtract 1 from 1 to get 0.

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

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

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

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

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

Combine 3x^{4} and 3x^{4} to get 6x^{4}.

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

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

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

Multiply -3 and -2 to get 6.

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

Combine 6x^{4} and -6x^{4} to get 0.

$$-2x^{3}$$

Expand

$$-2x^{3}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

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

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

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

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

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

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

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

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

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

To find the opposite of x^{6}+2x^{3}+1, find the opposite of each term.

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

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

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

Subtract 1 from 1 to get 0.

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

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

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

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

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

Combine 3x^{4} and 3x^{4} to get 6x^{4}.

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

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

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

Multiply -3 and -2 to get 6.

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

Combine 6x^{4} and -6x^{4} to get 0.

$$-2x^{3}$$