Solve 8x^6+999x^3=125

$$8 x ^ { 6 } + 999 x ^ { 3 } = 125$$

Solve for x

$$x=-5$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Subtract 125 from both sides.

$$8x^{6}+999x^{3}-125=0$$

Substitute t for x^{3}.

$$8t^{2}+999t-125=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 8 for a, 999 for b, and -125 for c in the quadratic formula.

$$t=\frac{-999±\sqrt{999^{2}-4\times 8\left(-125\right)}}{2\times 8}$$

Do the calculations.

$$t=\frac{-999±1001}{16}$$

Solve the equation t=\frac{-999±1001}{16} when ± is plus and when ± is minus.

$$t=\frac{1}{8} t=-125$$

Since x=t^{3}, the solutions are obtained by evaluating x=\sqrt[3]{t} for each t.

$$x=\frac{1}{2} x=-5$$