Lutasin ang f(x)=(x+2x^3)^4a=-1

$$f ( x ) = ( x + 2 x ^ { 3 } ) ^ { 4 } a = - 1$$

I-solve ang a

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

$$x\neq 0$$

Mga Hakbang sa Pagso-solve ng Linear Equation

Ang equation ay nasa standard form.

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

I-divide ang magkabilang dulo ng equation gamit ang \left(x+2x^{3}\right)^{4}.

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

Kapag na-divide gamit ang \left(x+2x^{3}\right)^{4}, ma-a-undo ang multiplication gamit ang \left(x+2x^{3}\right)^{4}.

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

I-divide ang -1 gamit ang \left(x+2x^{3}\right)^{4}.

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

I-solve ang x

$$x=\frac{\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}+\sqrt[3]{-6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}}{6}$$

$$x=\frac{\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}-54\sqrt[4]{-\frac{1}{a}}}-\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}}{6}\text{, }a<0$$