Question

Solve The Complex Number System0" alt="x^{3}-125=0" align="absmiddle" class="latex-formula">

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Answer 1

$\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2)~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x^3-125=0\implies x^3-5^3=0\implies (x-5)(x^2+5x+5^2)=0 \\\\[-0.35em] ~\dotfill\\\\ x-5=0\implies \boxed{x=5} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula}$

$\bf \stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{+25} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x=\cfrac{-5\pm\sqrt{5^2-4(1)(25)}}{2(1)}\implies x=\cfrac{-5\pm\sqrt{25-100}}{2} \\\\\\ x=\cfrac{-5\pm\sqrt{-75}}{2}~~ \begin{cases} 75=&3\cdot 5\cdot 5\\ &3\cdot 5^2 \end{cases}\implies x=\cfrac{-5\pm\sqrt{-3\cdot 5^2}}{2}$

$\bf x=\cfrac{-5\pm 5\sqrt{-3}}{2}\implies x=\cfrac{-5\pm 5\sqrt{-1\cdot 3}}{2}\implies x=\cfrac{-5\pm 5\sqrt{-1}\cdot \sqrt{3}}{2} \\\\\\ x=\cfrac{-5\pm 5i\sqrt{3}}{2}\implies \boxed{x= \begin{cases} \frac{-5+ 5i\sqrt{3}}{2}\\\\ \frac{-5-5i\sqrt{3}}{2} \end{cases}}$