Solve (1+x)^3=27

$$( 1 + x ) ^ { 3 } = 27$$

Solve for x (complex solution)

$$x=2$$

$$x=\frac{-5+3\sqrt{3}i}{2}\approx -2.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i-5}{2}\approx -2.5-2.598076211i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+x\right)^{3}.

$$1+3x+3x^{2}+x^{3}=27$$

Subtract 27 from both sides.

$$1+3x+3x^{2}+x^{3}-27=0$$

Subtract 27 from 1 to get -26.

$$-26+3x+3x^{2}+x^{3}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}+3x^{2}+3x-26=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x-26 by x-2 to get x^{2}+5x+13. Solve the equation where the result equals to 0.

$$x^{2}+5x+13=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 5 for b, and 13 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-27}}{2}$$

Solve the equation x^{2}+5x+13=0 when ± is plus and when ± is minus.

$$x=\frac{-3i\sqrt{3}-5}{2} x=\frac{-5+3i\sqrt{3}}{2}$$

List all found solutions.

$$x=2 x=\frac{-3i\sqrt{3}-5}{2} x=\frac{-5+3i\sqrt{3}}{2}$$

Solve for x

$$x=2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+x\right)^{3}.

$$1+3x+3x^{2}+x^{3}=27$$

Subtract 27 from both sides.

$$1+3x+3x^{2}+x^{3}-27=0$$

Subtract 27 from 1 to get -26.

$$-26+3x+3x^{2}+x^{3}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}+3x^{2}+3x-26=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x-26 by x-2 to get x^{2}+5x+13. Solve the equation where the result equals to 0.

$$x^{2}+5x+13=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 5 for b, and 13 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-27}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=2$$