How to solve (x - 1) ^ 3 = - 8

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x - 1) ^ 3 = - 8 | Equatix

Question

$$(x-1)^{3}=-8$$

Solve for x (complex solution)

$x=-1$
$x=2+\sqrt{3}i\approx 2+1.732050808i$
$x=-\sqrt{3}i+2\approx 2-1.732050808i$

Solution Steps

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

$$x^{3}-3x^{2}+3x-1=-8$$

Add $8$ to both sides.

$$x^{3}-3x^{2}+3x-1+8=0$$

Add $-1$ and $8$ to get $7$.

$$x^{3}-3x^{2}+3x+7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $7$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±7,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-3x^{2}+3x+7$ by $x+1$ to get $x^{2}-4x+7$. Solve the equation where the result equals to $0$.

$$x^{2}-4x+7=0$$

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 $1$ for $a$, $-4$ for $b$, and $7$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-12}}{2}$$

Solve the equation $x^{2}-4x+7=0$ when $±$ is plus and when $±$ is minus.

$$x=-\sqrt{3}i+2$$ $$x=2+\sqrt{3}i$$

List all found solutions.

$$x=-1$$ $$x=-\sqrt{3}i+2$$ $$x=2+\sqrt{3}i$$

Show Solution

Solve for x

$x=-1$

Solution Steps

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

$$x^{3}-3x^{2}+3x-1=-8$$

Add $8$ to both sides.

$$x^{3}-3x^{2}+3x-1+8=0$$

Add $-1$ and $8$ to get $7$.

$$x^{3}-3x^{2}+3x+7=0$$

$$±7,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-3x^{2}+3x+7$ by $x+1$ to get $x^{2}-4x+7$. Solve the equation where the result equals to $0$.

$$x^{2}-4x+7=0$$

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 $1$ for $a$, $-4$ for $b$, and $7$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-1$$

Show Solution

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Solve for x (complex solution)

Solution Steps

Solve for x

Solution Steps