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

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

Evaluate

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

Solution Steps

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}.

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

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

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

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

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

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

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

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

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

Subtract 1 from -1 to get -2.

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

Expand

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

Solution Steps

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}.

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

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

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

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

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

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

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

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

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

Subtract 1 from -1 to get -2.

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