Solve x^3=6^3+127

$$x ^ { 3 } = 6 ^ { 3 } + 127$$

Solve for x (complex solution)

$$x=\frac{-7\sqrt{3}i-7}{2}\approx -3.5-6.062177826i$$

$$x=7$$

$$x=\frac{-7+7\sqrt{3}i}{2}\approx -3.5+6.062177826i$$

Solution Steps

Calculate 6 to the power of 3 and get 216.

$$x^{3}=216+127$$

Add 216 and 127 to get 343.

$$x^{3}=343$$

Subtract 343 from both sides.

$$x^{3}-343=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -343 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±343,±49,±7,±1$$

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.

$$x=7$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-343 by x-7 to get x^{2}+7x+49. Solve the equation where the result equals to 0.

$$x^{2}+7x+49=0$$

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, 7 for b, and 49 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-7±\sqrt{-147}}{2}$$

Solve the equation x^{2}+7x+49=0 when ± is plus and when ± is minus.

$$x=\frac{-7i\sqrt{3}-7}{2} x=\frac{-7+7i\sqrt{3}}{2}$$

List all found solutions.

$$x=7 x=\frac{-7i\sqrt{3}-7}{2} x=\frac{-7+7i\sqrt{3}}{2}$$

Solve for x

$$x=7$$

Solution Steps

Calculate 6 to the power of 3 and get 216.

$$x^{3}=216+127$$

Add 216 and 127 to get 343.

$$x^{3}=343$$

Subtract 343 from both sides.

$$x^{3}-343=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -343 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±343,±49,±7,±1$$

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.

$$x=7$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-343 by x-7 to get x^{2}+7x+49. Solve the equation where the result equals to 0.

$$x^{2}+7x+49=0$$

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, 7 for b, and 49 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 49}}{2}$$

Do the calculations.

$$x=\frac{-7±\sqrt{-147}}{2}$$

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.

$$x=7$$