Solve left(x+3right)^3-x^3=279

$${ \left(x+3 \right) }^{ 3 } - { x }^{ 3 } =279$$

Solve for x

$$x=-7$$

$$x=4$$

Steps Using Factoring By Grouping

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

$$x^{3}+9x^{2}+27x+27-x^{3}=279$$

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

$$9x^{2}+27x+27=279$$

Subtract 279 from both sides.

$$9x^{2}+27x+27-279=0$$

Subtract 279 from 27 to get -252.

$$9x^{2}+27x-252=0$$

Divide both sides by 9.

$$x^{2}+3x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-28\right)=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite x^{2}+3x-28 as \left(x^{2}-4x\right)+\left(7x-28\right).

$$\left(x^{2}-4x\right)+\left(7x-28\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-4\right)+7\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+7\right)$$

To find equation solutions, solve x-4=0 and x+7=0.

$$x=4 x=-7$$

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+3\right)^{3}.

$$x^{3}+9x^{2}+27x+27-x^{3}=279$$

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

$$9x^{2}+27x+27=279$$

Subtract 279 from both sides.

$$9x^{2}+27x+27-279=0$$

Subtract 279 from 27 to get -252.

$$9x^{2}+27x-252=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 27 for b, and -252 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}-4\times 9\left(-252\right)}}{2\times 9}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\times 9\left(-252\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-27±\sqrt{729-36\left(-252\right)}}{2\times 9}$$

Multiply -36 times -252.

$$x=\frac{-27±\sqrt{729+9072}}{2\times 9}$$

Add 729 to 9072.

$$x=\frac{-27±\sqrt{9801}}{2\times 9}$$

Take the square root of 9801.

$$x=\frac{-27±99}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-27±99}{18}$$

Now solve the equation x=\frac{-27±99}{18} when ± is plus. Add -27 to 99.

$$x=\frac{72}{18}$$

Divide 72 by 18.

$$x=4$$

Now solve the equation x=\frac{-27±99}{18} when ± is minus. Subtract 99 from -27.

$$x=-\frac{126}{18}$$

Divide -126 by 18.

$$x=-7$$

The equation is now solved.

$$x=4 x=-7$$

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(x+3\right)^{3}.

$$x^{3}+9x^{2}+27x+27-x^{3}=279$$

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

$$9x^{2}+27x+27=279$$

Subtract 27 from both sides.

$$9x^{2}+27x=279-27$$

Subtract 27 from 279 to get 252.

$$9x^{2}+27x=252$$

Divide both sides by 9.

$$\frac{9x^{2}+27x}{9}=\frac{252}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{27}{9}x=\frac{252}{9}$$

Divide 27 by 9.

$$x^{2}+3x=\frac{252}{9}$$

Divide 252 by 9.

$$x^{2}+3x=28$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=28+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=28+\frac{9}{4}$$

Add 28 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{121}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{11}{2} x+\frac{3}{2}=-\frac{11}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=4 x=-7$$