How do you solve: #x^3/2=125#?

Question

2686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-05T19:32:47

#x=5root[3]2, \ 5\root[3]2e^{i{2\pi}/3},\ 5\root[3]2e^{i{4\pi}/3}#

#x^3/2=125#

#x^3=125\cdot 2#

#x^3=250#

#x^3=250e^{i0}#

#x^3=250e^{i2k\pi}#

#x=(250e^{i2k\pi})^{1/3}#

#x=\root[3]{250}e^{i{2k\pi}/3}#

#x=5\root[3]{2}e^{i{2k\pi}/3}#

Where, #k=0, 1, 2#

Now, setting the values of #k#, we get three roots of given cubic equation as follows

#x=5root[3]2, \ 5\root[3]2e^{i{2\pi}/3},\ 5\root[3]2e^{i{4\pi}/3}#