Solve (x+2)^3-(x-2)^3=1+(4x+1)(4x-1)

$$( x + 2 ) ^ { 3 } - ( x - 2 ) ^ { 3 } = 1 + ( 4 x + 1 ) ( 4 x - 1 )$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

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

$$x^{3}+6x^{2}+12x+8-\left(x-2\right)^{3}=1+\left(4x+1\right)\left(4x-1\right)$$

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

$$x^{3}+6x^{2}+12x+8-\left(x^{3}-6x^{2}+12x-8\right)=1+\left(4x+1\right)\left(4x-1\right)$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

$$x^{3}+6x^{2}+12x+8-x^{3}+6x^{2}-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

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

$$6x^{2}+12x+8+6x^{2}-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

Combine 6x^{2} and 6x^{2} to get 12x^{2}.

$$12x^{2}+12x+8-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

Combine 12x and -12x to get 0.

$$12x^{2}+8+8=1+\left(4x+1\right)\left(4x-1\right)$$

Add 8 and 8 to get 16.

$$12x^{2}+16=1+\left(4x+1\right)\left(4x-1\right)$$

Consider \left(4x+1\right)\left(4x-1\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.

$$12x^{2}+16=1+\left(4x\right)^{2}-1$$

Expand \left(4x\right)^{2}.

$$12x^{2}+16=1+4^{2}x^{2}-1$$

Calculate 4 to the power of 2 and get 16.

$$12x^{2}+16=1+16x^{2}-1$$

Subtract 1 from 1 to get 0.

$$12x^{2}+16=16x^{2}$$

Subtract 16x^{2} from both sides.

$$12x^{2}+16-16x^{2}=0$$

Combine 12x^{2} and -16x^{2} to get -4x^{2}.

$$-4x^{2}+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}=-16$$

Divide both sides by -4.

$$x^{2}=\frac{-16}{-4}$$

Divide -16 by -4 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

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

$$x^{3}+6x^{2}+12x+8-\left(x-2\right)^{3}=1+\left(4x+1\right)\left(4x-1\right)$$

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

$$x^{3}+6x^{2}+12x+8-\left(x^{3}-6x^{2}+12x-8\right)=1+\left(4x+1\right)\left(4x-1\right)$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

$$x^{3}+6x^{2}+12x+8-x^{3}+6x^{2}-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

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

$$6x^{2}+12x+8+6x^{2}-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

Combine 6x^{2} and 6x^{2} to get 12x^{2}.

$$12x^{2}+12x+8-12x+8=1+\left(4x+1\right)\left(4x-1\right)$$

Combine 12x and -12x to get 0.

$$12x^{2}+8+8=1+\left(4x+1\right)\left(4x-1\right)$$

Add 8 and 8 to get 16.

$$12x^{2}+16=1+\left(4x+1\right)\left(4x-1\right)$$

Consider \left(4x+1\right)\left(4x-1\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.

$$12x^{2}+16=1+\left(4x\right)^{2}-1$$

Expand \left(4x\right)^{2}.

$$12x^{2}+16=1+4^{2}x^{2}-1$$

Calculate 4 to the power of 2 and get 16.

$$12x^{2}+16=1+16x^{2}-1$$

Subtract 1 from 1 to get 0.

$$12x^{2}+16=16x^{2}$$

Subtract 16x^{2} from both sides.

$$12x^{2}+16-16x^{2}=0$$

Combine 12x^{2} and -16x^{2} to get -4x^{2}.

$$-4x^{2}+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 0 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 16}}{2\left(-4\right)}$$

Multiply 16 times 16.

$$x=\frac{0±\sqrt{256}}{2\left(-4\right)}$$

Take the square root of 256.

$$x=\frac{0±16}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±16}{-8}$$

Now solve the equation x=\frac{0±16}{-8} when ± is plus. Divide 16 by -8.

$$x=-2$$

Now solve the equation x=\frac{0±16}{-8} when ± is minus. Divide -16 by -8.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$