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 \(\frac{7- 7i\sqrt{3} }{10}\)
