What are the solutions to the equation (x^3+1)^3=8(2x-1) ?

The solutions to the equation (x^3+1)^3=8(2x-1) are x=1,x\approx 0.61803…,x\approx-1.61803…

Find the zeros of (x^3+1)^3=8(2x-1)

The zeros of (x^3+1)^3=8(2x-1) are x=1,x\approx 0.61803…,x\approx-1.61803…

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(x^{3} + 1)^{3} = 8 (2 x - 1)

x = 1, x \approx 0.61803 \dots, x \approx - 1.61803 \dots

Solution steps

(x^{3} + 1)^{3} = 8 (2 x - 1)

Expand

(x^{3} + 1)^{3} : x^{9} + 3 x^{6} + 3 x^{3} + 1

Expand

8 (2 x - 1) : 16 x - 8

x^{9} + 3 x^{6} + 3 x^{3} + 1 = 16 x - 8

Move

8

to the left side

x^{9} + 3 x^{6} + 3 x^{3} + 9 = 16 x

Move

16 x

to the left side

x^{9} + 3 x^{6} + 3 x^{3} + 9 - 16 x = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{9} + 3 x^{6} + 3 x^{3} - 16 x + 9 = 0

Factor

x^{9} + 3 x^{6} + 3 x^{3} - 16 x + 9 : (x - 1) (x^{8} + x^{7} + x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 7 x^{2} + 7 x - 9)

(x - 1) (x^{8} + x^{7} + x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 7 x^{2} + 7 x - 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{8} + x^{7} + x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 7 x^{2} + 7 x - 9 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{8} + x^{7} + x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 7 x^{2} + 7 x - 9 = 0 : x \approx 0.61803 \dots, x \approx - 1.61803 \dots

The solutions are

x = 1, x \approx 0.61803 \dots, x \approx - 1.61803 \dots

2 x^{3} + 5 x^{2} - 25 x = 0

A = (x + 1) (x - 1) (x^{2} + 1) + 1

c^{3} + 9 = 10

x^{4} + 4 x^{2} - 60 = 0

x + x^{3} = 2

What are the solutions to the equation (x^3+1)^3=8(2x-1) ?
The solutions to the equation (x^3+1)^3=8(2x-1) are x=1,x\approx 0.61803…,x\approx-1.61803…
Find the zeros of (x^3+1)^3=8(2x-1)
The zeros of (x^3+1)^3=8(2x-1) are x=1,x\approx 0.61803…,x\approx-1.61803…