Solve n(n^2-1)/6=4

$$\frac { n ( n ^ { 2 } - 1 ) } { 6 } = 4$$

Solve for n

$$n=3$$

Solution Steps

Multiply both sides by 6.

$$n\left(n^{2}-1\right)=4\times 6$$

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

$$n^{3}-n=4\times 6$$

Multiply 4 and 6 to get 24.

$$n^{3}-n=24$$

Subtract 24 from both sides.

$$n^{3}-n-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$n=3$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}-n-24 by n-3 to get n^{2}+3n+8. Solve the equation where the result equals to 0.

$$n^{2}+3n+8=0$$

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, 3 for b, and 8 for c in the quadratic formula.

$$n=\frac{-3±\sqrt{3^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$n=\frac{-3±\sqrt{-23}}{2}$$

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.

$$n=3$$