Solve 16(x^3+1)^2-22(x^3+1)-3=0

$$16 ( x ^ { 3 } + 1 ) ^ { 2 } - 22 ( x ^ { 3 } + 1 ) - 3 = 0$$

Solve for x (complex solution)

$$x\in \frac{2^{\frac{2}{3}}}{2},\frac{2^{\frac{2}{3}}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}e^{\frac{2\pi i}{3}}}{2},\frac{3^{\frac{2}{3}}e^{\frac{5\pi i}{3}}}{2},\frac{3^{\frac{2}{3}}e^{\frac{\pi i}{3}}}{2},-\frac{3^{\frac{2}{3}}}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{3}+1\right)^{2}.

$$16\left(\left(x^{3}\right)^{2}+2x^{3}+1\right)-22\left(x^{3}+1\right)-3=0$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$16\left(x^{6}+2x^{3}+1\right)-22\left(x^{3}+1\right)-3=0$$

Use the distributive property to multiply 16 by x^{6}+2x^{3}+1.

$$16x^{6}+32x^{3}+16-22\left(x^{3}+1\right)-3=0$$

Use the distributive property to multiply -22 by x^{3}+1.

$$16x^{6}+32x^{3}+16-22x^{3}-22-3=0$$

Combine 32x^{3} and -22x^{3} to get 10x^{3}.

$$16x^{6}+10x^{3}+16-22-3=0$$

Subtract 22 from 16 to get -6.

$$16x^{6}+10x^{3}-6-3=0$$

Subtract 3 from -6 to get -9.

$$16x^{6}+10x^{3}-9=0$$

Substitute t for x^{3}.

$$16t^{2}+10t-9=0$$

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, 10 for b, and -9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\times 16\left(-9\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-10±26}{32}$$

Solve the equation t=\frac{-10±26}{32} when ± is plus and when ± is minus.

$$t=\frac{1}{2} t=-\frac{9}{8}$$

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

$$x=-\frac{e^{\frac{\pi i}{3}}}{\sqrt[3]{2}} x=\frac{ie^{\frac{\pi i}{6}}}{\sqrt[3]{2}} x=\frac{1}{\sqrt[3]{2}} x=-\frac{\sqrt[3]{9}ie^{\frac{\pi i}{6}}}{2} x=-\frac{\sqrt[3]{9}}{2} x=\frac{\sqrt[3]{9}e^{\frac{\pi i}{3}}}{2}$$

Solve for x

$$x = -\frac{3 ^ {\frac{2}{3}}}{2} \approx -1.040041912$$

$$x=\frac{2^{\frac{2}{3}}}{2}\approx 0.793700526$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{3}+1\right)^{2}.

$$16\left(\left(x^{3}\right)^{2}+2x^{3}+1\right)-22\left(x^{3}+1\right)-3=0$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$16\left(x^{6}+2x^{3}+1\right)-22\left(x^{3}+1\right)-3=0$$

Use the distributive property to multiply 16 by x^{6}+2x^{3}+1.

$$16x^{6}+32x^{3}+16-22\left(x^{3}+1\right)-3=0$$

Use the distributive property to multiply -22 by x^{3}+1.

$$16x^{6}+32x^{3}+16-22x^{3}-22-3=0$$

Combine 32x^{3} and -22x^{3} to get 10x^{3}.

$$16x^{6}+10x^{3}+16-22-3=0$$

Subtract 22 from 16 to get -6.

$$16x^{6}+10x^{3}-6-3=0$$

Subtract 3 from -6 to get -9.

$$16x^{6}+10x^{3}-9=0$$

Substitute t for x^{3}.

$$16t^{2}+10t-9=0$$

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, 10 for b, and -9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\times 16\left(-9\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-10±26}{32}$$

Solve the equation t=\frac{-10±26}{32} when ± is plus and when ± is minus.

$$t=\frac{1}{2} t=-\frac{9}{8}$$

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

$$x=\frac{1}{\sqrt[3]{2}} x=-\frac{\sqrt[3]{9}}{2}$$