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Solve for x (complex solution)
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Solve for x
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x^{3}-216=0
Subtract 216 from both sides.
±216,±108,±72,±54,±36,±27,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=6
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}+6x+36=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-216 by x-6 to get x^{2}+6x+36. Solve the equation where the result equals to 0.
x=\frac{-6±\sqrt{6^{2}-4\times 1\times 36}}{2}
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 36 for c in the quadratic formula.
x=\frac{-6±\sqrt{-108}}{2}
Do the calculations.
x=-3i\sqrt{3}-3 x=-3+3i\sqrt{3}
Solve the equation x^{2}+6x+36=0 when ± is plus and when ± is minus.
x=6 x=-3i\sqrt{3}-3 x=-3+3i\sqrt{3}
List all found solutions.
x^{3}-216=0
Subtract 216 from both sides.
±216,±108,±72,±54,±36,±27,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=6
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}+6x+36=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-216 by x-6 to get x^{2}+6x+36. Solve the equation where the result equals to 0.
x=\frac{-6±\sqrt{6^{2}-4\times 1\times 36}}{2}
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 36 for c in the quadratic formula.
x=\frac{-6±\sqrt{-108}}{2}
Do the calculations.
x\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x=6
List all found solutions.