Solve 2/n^3+n^2=1

$$\frac { 2 } { n ^ { 3 } + n ^ { 2 } } = 1$$

Solve for n (complex solution)

$$n=-1-i$$

$$n=1$$

$$n=-1+i$$

Solution Steps

Variable n cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by \left(n+1\right)n^{2}.

$$2=\left(n+1\right)n^{2}$$

Use the distributive property to multiply n+1 by n^{2}.

$$2=n^{3}+n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$n^{3}+n^{2}=2$$

Subtract 2 from both sides.

$$n^{3}+n^{2}-2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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.

$$n=1$$

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

$$n^{2}+2n+2=0$$

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, 2 for b, and 2 for c in the quadratic formula.

$$n=\frac{-2±\sqrt{2^{2}-4\times 1\times 2}}{2}$$

Do the calculations.

$$n=\frac{-2±\sqrt{-4}}{2}$$

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

$$n=-1-i n=-1+i$$

List all found solutions.

$$n=1 n=-1-i n=-1+i$$

Solve for n

$$n=1$$

Solution Steps

Variable n cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by \left(n+1\right)n^{2}.

$$2=\left(n+1\right)n^{2}$$

Use the distributive property to multiply n+1 by n^{2}.

$$2=n^{3}+n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$n^{3}+n^{2}=2$$

Subtract 2 from both sides.

$$n^{3}+n^{2}-2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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.

$$n=1$$

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

$$n^{2}+2n+2=0$$

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, 2 for b, and 2 for c in the quadratic formula.

$$n=\frac{-2±\sqrt{2^{2}-4\times 1\times 2}}{2}$$

Do the calculations.

$$n=\frac{-2±\sqrt{-4}}{2}$$

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

$$n\in \emptyset$$

List all found solutions.

$$n=1$$