Solve left(x+1right)^3/4=16

$$\frac{ { \left(x+1 \right) }^{ 3 } }{ 4 } =16$$

Solve for x (complex solution)

$$x=3$$

$$x=-3+2\sqrt{3}i\approx -3+3.464101615i$$

$$x=-2\sqrt{3}i-3\approx -3-3.464101615i$$

Solution Steps

Multiply both sides by 4.

$$\left(x+1\right)^{3}=16\times 4$$

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

$$x^{3}+3x^{2}+3x+1=16\times 4$$

Multiply 16 and 4 to get 64.

$$x^{3}+3x^{2}+3x+1=64$$

Subtract 64 from both sides.

$$x^{3}+3x^{2}+3x+1-64=0$$

Subtract 64 from 1 to get -63.

$$x^{3}+3x^{2}+3x-63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -63 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±63,±21,±9,±7,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x-63 by x-3 to get x^{2}+6x+21. Solve the equation where the result equals to 0.

$$x^{2}+6x+21=0$$

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, 6 for b, and 21 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 21}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-48}}{2}$$

Solve the equation x^{2}+6x+21=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=3$$

Solution Steps

Multiply both sides by 4.

$$\left(x+1\right)^{3}=16\times 4$$

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

$$x^{3}+3x^{2}+3x+1=16\times 4$$

Multiply 16 and 4 to get 64.

$$x^{3}+3x^{2}+3x+1=64$$

Subtract 64 from both sides.

$$x^{3}+3x^{2}+3x+1-64=0$$

Subtract 64 from 1 to get -63.

$$x^{3}+3x^{2}+3x-63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -63 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±63,±21,±9,±7,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}+3x-63 by x-3 to get x^{2}+6x+21. Solve the equation where the result equals to 0.

$$x^{2}+6x+21=0$$

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, 6 for b, and 21 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 21}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-48}}{2}$$

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.

$$x=3$$