Solve left(x^3+1right)^2+(2x-3)(x^3-1)+4

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

Evaluate

$$x^{6}+2x^{4}-x^{3}-2x+8$$

Solution Steps

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

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

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

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

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

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

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

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

Add 1 and 3 to get 4.

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

Add 4 and 4 to get 8.

$$x^{6}-x^{3}+8+2x^{4}-2x$$

Expand

$$x^{6}+2x^{4}-x^{3}-2x+8$$

Solution Steps

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

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

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

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

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

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

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

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

Add 1 and 3 to get 4.

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

Add 4 and 4 to get 8.

$$x^{6}-x^{3}+8+2x^{4}-2x$$