What are the solutions to the equation (3(n+1)n(n-1))/6 =(n(n-1))/2 ?

The solutions to the equation (3(n+1)n(n-1))/6 =(n(n-1))/2 are n=1,n=0

Find the zeros of (3(n+1)n(n-1))/6 =(n(n-1))/2

The zeros of (3(n+1)n(n-1))/6 =(n(n-1))/2 are n=1,n=0

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\frac{3 ( n + 1 ) n ( n - 1 )}{6} = \frac{n ( n - 1 )}{2}

n = 1, n = 0

Solution steps

\frac{3 ( n + 1 ) n ( n - 1 )}{6} = \frac{n ( n - 1 )}{2}

Find Least Common Multiplier of

6, 2 : 6

Multiply by LCM=

6

\frac{3 ( n + 1 ) n ( n - 1 )}{6} \cdot 6 = \frac{n ( n - 1 )}{2} \cdot 6

Simplify

3 n (n + 1) (n - 1) = 3 n (n - 1)

Expand

3 n (n + 1) (n - 1) : 3 n^{3} - 3 n

Expand

3 n (n - 1) : 3 n^{2} - 3 n

3 n^{3} - 3 n = 3 n^{2} - 3 n

Move

3 n

to the left side

3 n^{3} = 3 n^{2}

Move

3 n^{2}

to the left side

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

Find one solution for

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

using Newton-Raphson:

n \approx 1.00000 \dots

Apply long division:

\frac{3 n ^{3} - 3 n ^{2}}{n - 1} = 3 n^{2}

3 n^{2} \approx 0

Find one solution for

3 n^{2} = 0

using Newton-Raphson:

n \approx 0.00000 \dots

Apply long division:

\frac{3 n ^{2}}{n} = 3 n

3 n \approx 0

n \approx 0

Approximate to algebraic form

n = 0

Approximate to algebraic form

n \approx 1.00000 \dots : n = 1

n = 1, n = 0

The solutions are

n = 1, n = 0

(x + 6)^{3} = - 27

z^{4} + 4 z^{2} + 16 = 0

x^{4} - 29 x^{2} + 100 = 0

z^{3} = 27

x^{4} - 4 x^{3} + x^{2} + 4 x + 1 = 0

What are the solutions to the equation (3(n+1)n(n-1))/6 =(n(n-1))/2 ?
The solutions to the equation (3(n+1)n(n-1))/6 =(n(n-1))/2 are n=1,n=0
Find the zeros of (3(n+1)n(n-1))/6 =(n(n-1))/2
The zeros of (3(n+1)n(n-1))/6 =(n(n-1))/2 are n=1,n=0