Solve A=(3x^2-x^2)^4

$$A = ( 3 x ^ { 2 } - x ^ { 2 } ) ^ { 4 }$$

Solve for A

$$A=16x^{8}$$

Solution Steps

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$A=\left(2x^{2}\right)^{4}$$

Expand \left(2x^{2}\right)^{4}.

$$A=2^{4}\left(x^{2}\right)^{4}$$

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

$$A=2^{4}x^{8}$$

Calculate 2 to the power of 4 and get 16.

$$A=16x^{8}$$

Solve for x (complex solution)

$$x\in \left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[8]{A},\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[8]{A},-\frac{\sqrt{2}i\sqrt[8]{A}}{2},\frac{\sqrt{2}i\sqrt[8]{A}}{2},-\frac{\sqrt{2}\sqrt[8]{A}}{2},\frac{\sqrt{2}\sqrt[8]{A}}{2},\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[8]{A},\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[8]{A}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{2}\sqrt[8]{A}}{2}\text{; }x=\frac{\sqrt{2}\sqrt[8]{A}}{2}\text{, }&A\geq 0\\x=0\text{, }&A=0\end{matrix}\right.$$