Solve left(x+3right)^3=8

$${ \left(x+3 \right) }^{ 3 } =8$$

Solve for x (complex solution)

$$x=-1$$

$$x=-4+\sqrt{3}i\approx -4+1.732050808i$$

$$x=-\sqrt{3}i-4\approx -4-1.732050808i$$

Solution Steps

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

$$x^{3}+9x^{2}+27x+27=8$$

Subtract 8 from both sides.

$$x^{3}+9x^{2}+27x+27-8=0$$

Subtract 8 from 27 to get 19.

$$x^{3}+9x^{2}+27x+19=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 19 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±19,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+9x^{2}+27x+19 by x+1 to get x^{2}+8x+19. Solve the equation where the result equals to 0.

$$x^{2}+8x+19=0$$

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, 8 for b, and 19 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-12}}{2}$$

Solve the equation x^{2}+8x+19=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3}i-4 x=-4+\sqrt{3}i$$

List all found solutions.

$$x=-1 x=-\sqrt{3}i-4 x=-4+\sqrt{3}i$$

Solve for x

$$x=-1$$

Solution Steps

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

$$x^{3}+9x^{2}+27x+27=8$$

Subtract 8 from both sides.

$$x^{3}+9x^{2}+27x+27-8=0$$

Subtract 8 from 27 to get 19.

$$x^{3}+9x^{2}+27x+19=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 19 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±19,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+9x^{2}+27x+19 by x+1 to get x^{2}+8x+19. Solve the equation where the result equals to 0.

$$x^{2}+8x+19=0$$

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, 8 for b, and 19 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$x=\frac{-8±\sqrt{-12}}{2}$$

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.

$$x=-1$$