Question

How can I find the complex roots of 125x^3+343. Please explain your steps

Answer 1

First we assume that this is equal to zero

so what we notice is this is a sum of 2 perfect cube

recall that
a^3+b^3=(a+b)(a^2-ab+b^2)

125x^3+343=
(5x)³+(7)³=(5x+7)(25x²-35x+49)

now solve the part in second parenthasees because we can see that the first parenthasees (5x+7) is going to have a real root

remember quadratic formula

for ax^2+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
a=25
b=-35
c=49

x=$\frac{-(-35)+/- \sqrt{(-35)^2-4(25)(49)} }{2(25)}$
x=$\frac{35+/- \sqrt{1225-4900} }{50}$
x=$\frac{35+/- \sqrt{-3675} }{50}$
x=$\frac{35+/- (\sqrt{3675})( \sqrt{-1}) }{50}$
x=$\frac{35+/- (\sqrt{3675})(i) }{50}$
x=$\frac{35+/- 35i\sqrt{3} }{50}$
x=$\frac{7+/- 7i\sqrt{3} }{10}$

the comlex roots are

x=$\frac{7+ 7i\sqrt{3} }{10}$ and $\frac{7- 7i\sqrt{3} }{10}$