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

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

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

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

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

n = 0, n = 1, n = - 1

Solution steps

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

Multiply both sides by

3

3 n (n + 1) = n (n + 1) (n + 2)

Expand

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

Expand

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

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

Switch sides

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

Move

3 n

to the left side

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

Move

3 n^{2}

to the left side

n^{3} - n = 0

Find one solution for

n^{3} - n = 0

using Newton-Raphson:

n \approx 0.00000 \dots

Apply long division:

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

n^{2} - 1 \approx 0

Find one solution for

n^{2} - 1 = 0

using Newton-Raphson:

n \approx 1.00000 \dots

Apply long division:

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

n + 1 \approx 0

n \approx - 1

Approximate to algebraic form

n = - 1

Approximate to algebraic form

n \approx 0.00000 \dots : n = 0

n \approx 1.00000 \dots : n = 1

n = 0, n = 1, n = - 1

The solutions are

n = 0, n = 1, n = - 1

- 64 = m^{3}

x, 0 = x^{3} - 2 x + 5

3500 \cdot x^{10} = 7000

(1 + 0.01)^{12} = (1 + x)^{6}

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

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