Solve y=(x^4-1)^3

$$y = ( x ^ { 4 } - 1 ) ^ { 3 }$$

Solve for y (complex solution)

$$y=x^{12}-3x^{8}+3x^{4}-1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x^{4}-1\right)^{3}.

$$y=\left(x^{4}\right)^{3}-3\left(x^{4}\right)^{2}+3x^{4}-1$$

To raise a power to another power, multiply the exponents. Multiply 4 and 3 to get 12.

$$y=x^{12}-3\left(x^{4}\right)^{2}+3x^{4}-1$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$y=x^{12}-3x^{8}+3x^{4}-1$$

Solve for y

$$y=\left(x^{4}-1\right)^{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x^{4}-1\right)^{3}.

$$y=\left(x^{4}\right)^{3}-3\left(x^{4}\right)^{2}+3x^{4}-1$$

To raise a power to another power, multiply the exponents. Multiply 4 and 3 to get 12.

$$y=x^{12}-3\left(x^{4}\right)^{2}+3x^{4}-1$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$y=x^{12}-3x^{8}+3x^{4}-1$$

Solve for x (complex solution)

$$x\in \sqrt[4]{e^{\frac{4\pi i}{3}}\sqrt[3]{y}+1},-i\sqrt[4]{e^{\frac{4\pi i}{3}}\sqrt[3]{y}+1},i\sqrt[4]{e^{\frac{4\pi i}{3}}\sqrt[3]{y}+1},-\sqrt[4]{e^{\frac{4\pi i}{3}}\sqrt[3]{y}+1},-i\sqrt[4]{\sqrt[3]{y}+1},\sqrt[4]{\sqrt[3]{y}+1},i\sqrt[4]{\sqrt[3]{y}+1},-\sqrt[4]{\sqrt[3]{y}+1},-i\sqrt[4]{e^{\frac{2\pi i}{3}}\sqrt[3]{y}+1},\sqrt[4]{e^{\frac{2\pi i}{3}}\sqrt[3]{y}+1},i\sqrt[4]{e^{\frac{2\pi i}{3}}\sqrt[3]{y}+1},-\sqrt[4]{e^{\frac{2\pi i}{3}}\sqrt[3]{y}+1}$$

Solve for x

$$x=\sqrt[4]{\sqrt[3]{y}+1}$$

$$x=-\sqrt[4]{\sqrt[3]{y}+1}\text{, }y\geq -1$$