Solve (x+2)^3-x(x-3)(x+3)=-16+6(x+1)^2

$$( x + 2 ) ^ { 3 } - x ( x - 3 ) ( x + 3 ) = - 16 + 6 ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=-2$$

Steps for Solving Linear Equation

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

$$x^{3}+6x^{2}+12x+8-x\left(x-3\right)\left(x+3\right)=-16+6\left(x+1\right)^{2}$$

Use the distributive property to multiply x by x-3.

$$x^{3}+6x^{2}+12x+8-\left(x^{2}-3x\right)\left(x+3\right)=-16+6\left(x+1\right)^{2}$$

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

$$x^{3}+6x^{2}+12x+8-\left(x^{3}-9x\right)=-16+6\left(x+1\right)^{2}$$

To find the opposite of x^{3}-9x, find the opposite of each term.

$$x^{3}+6x^{2}+12x+8-x^{3}+9x=-16+6\left(x+1\right)^{2}$$

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

$$6x^{2}+12x+8+9x=-16+6\left(x+1\right)^{2}$$

Combine 12x and 9x to get 21x.

$$6x^{2}+21x+8=-16+6\left(x+1\right)^{2}$$

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

$$6x^{2}+21x+8=-16+6\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply 6 by x^{2}+2x+1.

$$6x^{2}+21x+8=-16+6x^{2}+12x+6$$

Add -16 and 6 to get -10.

$$6x^{2}+21x+8=-10+6x^{2}+12x$$

Subtract 6x^{2} from both sides.

$$6x^{2}+21x+8-6x^{2}=-10+12x$$

Combine 6x^{2} and -6x^{2} to get 0.

$$21x+8=-10+12x$$

Subtract 12x from both sides.

$$21x+8-12x=-10$$

Combine 21x and -12x to get 9x.

$$9x+8=-10$$

Subtract 8 from both sides.

$$9x=-10-8$$

Subtract 8 from -10 to get -18.

$$9x=-18$$

Divide both sides by 9.

$$x=\frac{-18}{9}$$

Divide -18 by 9 to get -2.

$$x=-2$$