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

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

Expand

$$x^{3}+8x^{2}+33x+18$$

Solution Steps

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

$$x^{3}+9x^{2}+27x+27-\left(3-x\right)^{2}$$

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

$$x^{3}+9x^{2}+27x+27-\left(9-6x+x^{2}\right)$$

To find the opposite of 9-6x+x^{2}, find the opposite of each term.

$$x^{3}+9x^{2}+27x+27-9+6x-x^{2}$$

Subtract 9 from 27 to get 18.

$$x^{3}+9x^{2}+27x+18+6x-x^{2}$$

Combine 27x and 6x to get 33x.

$$x^{3}+9x^{2}+33x+18-x^{2}$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$x^{3}+8x^{2}+33x+18$$

Evaluate

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

Solution Steps

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

$$x^{3}+9x^{2}+27x+27-\left(3-x\right)^{2}$$

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

$$x^{3}+9x^{2}+27x+27-\left(9-6x+x^{2}\right)$$

To find the opposite of 9-6x+x^{2}, find the opposite of each term.

$$x^{3}+9x^{2}+27x+27-9+6x-x^{2}$$

Subtract 9 from 27 to get 18.

$$x^{3}+9x^{2}+27x+18+6x-x^{2}$$

Combine 27x and 6x to get 33x.

$$x^{3}+9x^{2}+33x+18-x^{2}$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$x^{3}+8x^{2}+33x+18$$