Solve (k+2)^3-k^3=20

$$( k + 2 ) ^ { 3 } - k ^ { 3 } = 20$$

Solve for k

$$k=\sqrt{3}-1\approx 0.732050808$$

$$k=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

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(k+2\right)^{3}.

$$k^{3}+6k^{2}+12k+8-k^{3}=20$$

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

$$6k^{2}+12k+8=20$$

Subtract 20 from both sides.

$$6k^{2}+12k+8-20=0$$

Subtract 20 from 8 to get -12.

$$6k^{2}+12k-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 12 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-12±\sqrt{12^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 12.

$$k=\frac{-12±\sqrt{144-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$k=\frac{-12±\sqrt{144-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$k=\frac{-12±\sqrt{144+288}}{2\times 6}$$

Add 144 to 288.

$$k=\frac{-12±\sqrt{432}}{2\times 6}$$

Take the square root of 432.

$$k=\frac{-12±12\sqrt{3}}{2\times 6}$$

Multiply 2 times 6.

$$k=\frac{-12±12\sqrt{3}}{12}$$

Now solve the equation k=\frac{-12±12\sqrt{3}}{12} when ± is plus. Add -12 to 12\sqrt{3}.

$$k=\frac{12\sqrt{3}-12}{12}$$

Divide -12+12\sqrt{3} by 12.

$$k=\sqrt{3}-1$$

Now solve the equation k=\frac{-12±12\sqrt{3}}{12} when ± is minus. Subtract 12\sqrt{3} from -12.

$$k=\frac{-12\sqrt{3}-12}{12}$$

Divide -12-12\sqrt{3} by 12.

$$k=-\sqrt{3}-1$$

The equation is now solved.

$$k=\sqrt{3}-1 k=-\sqrt{3}-1$$

Steps for Completing the Square

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

$$k^{3}+6k^{2}+12k+8-k^{3}=20$$

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

$$6k^{2}+12k+8=20$$

Subtract 8 from both sides.

$$6k^{2}+12k=20-8$$

Subtract 8 from 20 to get 12.

$$6k^{2}+12k=12$$

Divide both sides by 6.

$$\frac{6k^{2}+12k}{6}=\frac{12}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$k^{2}+\frac{12}{6}k=\frac{12}{6}$$

Divide 12 by 6.

$$k^{2}+2k=\frac{12}{6}$$

Divide 12 by 6.

$$k^{2}+2k=2$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+2k+1^{2}=2+1^{2}$$

Square 1.

$$k^{2}+2k+1=2+1$$

Add 2 to 1.

$$k^{2}+2k+1=3$$

Factor k^{2}+2k+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+1\right)^{2}}=\sqrt{3}$$

Simplify.

$$k+1=\sqrt{3} k+1=-\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$k=\sqrt{3}-1 k=-\sqrt{3}-1$$