Solve a^3-1/a-1=-3a^2

$$\frac { a ^ { 3 } - 1 } { a - 1 } = - 3 a ^ { 2 }$$

Solve for a

$$a=\frac{-1+\sqrt{15}i}{8}\approx -0.125+0.484122918i$$

$$a=\frac{-\sqrt{15}i-1}{8}\approx -0.125-0.484122918i$$

Solution Steps

Variable a cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by a-1.

$$a^{3}-1=-3a^{2}\left(a-1\right)$$

Use the distributive property to multiply -3a^{2} by a-1.

$$a^{3}-1=-3a^{3}+3a^{2}$$

Add 3a^{3} to both sides.

$$a^{3}-1+3a^{3}=3a^{2}$$

Combine a^{3} and 3a^{3} to get 4a^{3}.

$$4a^{3}-1=3a^{2}$$

Subtract 3a^{2} from both sides.

$$4a^{3}-1-3a^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$4a^{3}-3a^{2}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide 4a^{3}-3a^{2}-1 by a-1 to get 4a^{2}+a+1. Solve the equation where the result equals to 0.

$$4a^{2}+a+1=0$$

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 4 for a, 1 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{-1±\sqrt{-15}}{8}$$

Solve the equation 4a^{2}+a+1=0 when ± is plus and when ± is minus.

$$a=\frac{-\sqrt{15}i-1}{8} a=\frac{-1+\sqrt{15}i}{8}$$

Remove the values that the variable cannot be equal to.

$$a\in \emptyset$$

List all found solutions.

$$a=1 a=\frac{-\sqrt{15}i-1}{8} a=\frac{-1+\sqrt{15}i}{8}$$

Variable a cannot be equal to 1.

$$a=\frac{-1+\sqrt{15}i}{8} a=\frac{-\sqrt{15}i-1}{8}$$