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

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

Evaluate

$$-3x^{4}$$

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}.

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

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

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

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

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

Consider \left(1-x^{3}\right)\left(1+x^{3}\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 1.

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

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

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

Add -1 and 1 to get 0.

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

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

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

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

$$-3x^{4}$$

Expand

$$-3x^{4}$$

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}.

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

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

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

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

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

Consider \left(1-x^{3}\right)\left(1+x^{3}\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 1.

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

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

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

Add -1 and 1 to get 0.

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

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

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

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

$$-3x^{4}$$