Solve 16x^6=4x^3+12x^2

$$16 x ^ { 6 } = 4 x ^ { 3 } + 12 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt[3]{9\sqrt{41}-53}-\sqrt[3]{9\sqrt{41}+53}-2}{6}\approx -0.85567287$$

$$x=1$$

$$x=\frac{\left(1+\sqrt{3}i\right)\left(2\sqrt[3]{9\sqrt{41}+53}+\sqrt[3]{9\sqrt{41}-53}-2+\sqrt{3}\sqrt[3]{9\sqrt{41}-53}i+2\sqrt{3}i\right)}{24}\approx -0.072163565+0.933432167i$$

$$x=\frac{\left(-\sqrt{3}i+1\right)\left(-\sqrt{3}\sqrt[3]{9\sqrt{41}-53}i+2\sqrt[3]{9\sqrt{41}+53}+\sqrt[3]{9\sqrt{41}-53}-2\sqrt{3}i-2\right)}{24}\approx -0.072163565-0.933432167i$$

$$x=0$$

Steps Using the Quadratic Formula

Subtract 4x^{3} from both sides.

$$16x^{6}-4x^{3}=12x^{2}$$

Subtract 12x^{2} from both sides.

$$16x^{6}-4x^{3}-12x^{2}=0$$

Substitute t for x^{3}.

$$16t^{2}-4t-12=0$$

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 16 for a, -4 for b, and -12 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 16\left(-12\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{4±28}{32}$$

Solve the equation t=\frac{4±28}{32} when ± is plus and when ± is minus.

$$t=1 t=-\frac{3}{4}$$

Since x=t^{3}, the solutions are obtained by solving the equation for each t.

$$x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2} x=1 x=-\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}ie^{\frac{\pi i}{6}}}{4} x=-\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}}{4} x=\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}e^{\frac{\pi i}{3}}}{4}$$