Solve a^3-1/a-1=21

$$\frac { a ^ { 3 } - 1 } { a - 1 } = 21$$

Solve for a

$$a=-5$$

$$a=4$$

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.

$$a^{3}-1=21\left(a-1\right)$$

Use the distributive property to multiply 21 by a-1.

$$a^{3}-1=21a-21$$

Subtract 21a from both sides.

$$a^{3}-1-21a=-21$$

Add 21 to both sides.

$$a^{3}-1-21a+21=0$$

Add -1 and 21 to get 20.

$$a^{3}+20-21a=0$$

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

$$a^{3}-21a+20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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 a^{3}-21a+20 by a-1 to get a^{2}+a-20. Solve the equation where the result equals to 0.

$$a^{2}+a-20=0$$

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, 1 for b, and -20 for c in the quadratic formula.

$$a=\frac{-1±\sqrt{1^{2}-4\times 1\left(-20\right)}}{2}$$

Do the calculations.

$$a=\frac{-1±9}{2}$$

Solve the equation a^{2}+a-20=0 when ± is plus and when ± is minus.

$$a=-5 a=4$$

Remove the values that the variable cannot be equal to.

$$a=4\text{ or }a=-5$$

List all found solutions.

$$a=1 a=-5 a=4$$

Variable a cannot be equal to 1.

$$a=4 a=-5$$