Solve {l}{x^{2}(x-4)=x^{(2+1)}-4=x^3-}{3x^3(3x^3-9x^2+x-10)=}

$$\left. \begin{array} { l } { x ^ { 2 } ( x - 4 ) = x ^ { ( 2 + 1 ) } - 4 = x ^ { 3 } - } \\ { 3 x ^ { 3 } ( 3 x ^ { 3 } - 9 x ^ { 2 } + x - 10 ) = } \end{array} \right.$$

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

Use the distributive property to multiply x^{2} by x-4.

$$x^{3}-4x^{2}=x^{2+1}-4$$

Add 2 and 1 to get 3.

$$x^{3}-4x^{2}=x^{3}-4$$

Subtract x^{3} from both sides.

$$x^{3}-4x^{2}-x^{3}=-4$$

Combine x^{3} and -x^{3} to get 0.

$$-4x^{2}=-4$$

Divide both sides by -4.

$$x^{2}=\frac{-4}{-4}$$

Divide -4 by -4 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x^{2} by x-4.

$$x^{3}-4x^{2}=x^{2+1}-4$$

Add 2 and 1 to get 3.

$$x^{3}-4x^{2}=x^{3}-4$$

Subtract x^{3} from both sides.

$$x^{3}-4x^{2}-x^{3}=-4$$

Combine x^{3} and -x^{3} to get 0.

$$-4x^{2}=-4$$

Add 4 to both sides.

$$-4x^{2}+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 0 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 4}}{2\left(-4\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 4}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{0±\sqrt{64}}{2\left(-4\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±8}{-8}$$

Now solve the equation x=\frac{0±8}{-8} when ± is plus. Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{0±8}{-8} when ± is minus. Divide -8 by -8.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$