Solve x ^ 3 - x ^ 2 = (x - 1) ^ 3 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x^{3}-x^{2}=(x-1)^{3}$$

Solve for x

$x=\frac{1}{2}=0.5$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

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

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

Subtract $x^{3}$ from both sides.

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

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

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

Add $3x^{2}$ to both sides.

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

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

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

Subtract $3x$ from both sides.

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

Add $1$ to both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-3$$ $$ab=2\times 1=2$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.

$$a=-2$$ $$b=-1$$

Rewrite $2x^{2}-3x+1$ as $\left(2x^{2}-2x\right)+\left(-x+1\right)$.

$$\left(2x^{2}-2x\right)+\left(-x+1\right)$$

Factor out $2x$ in the first and $-1$ in the second group.

$$2x\left(x-1\right)-\left(x-1\right)$$

Factor out common term $x-1$ by using distributive property.

$$\left(x-1\right)\left(2x-1\right)$$

To find equation solutions, solve $x-1=0$ and $2x-1=0$.

$$x=1$$ $$x=\frac{1}{2}$$

Steps Using the Quadratic Formula

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

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

Subtract $x^{3}$ from both sides.

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

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

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

Add $3x^{2}$ to both sides.

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

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

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

Subtract $3x$ from both sides.

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

Add $1$ to both sides.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-3$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2}}{2\times 2}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Add $9$ to $-8$.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\times 2}$$

Take the square root of $1$.

$$x=\frac{-\left(-3\right)±1}{2\times 2}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±1}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{3±1}{4}$$

Now solve the equation $x=\frac{3±1}{4}$ when $±$ is plus. Add $3$ to $1$.

$$x=\frac{4}{4}$$

Divide $4$ by $4$.

$$x=1$$

Now solve the equation $x=\frac{3±1}{4}$ when $±$ is minus. Subtract $1$ from $3$.

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

Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=1$$ $$x=\frac{1}{2}$$

Steps for Completing the Square

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

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

Subtract $x^{3}$ from both sides.

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

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

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

Add $3x^{2}$ to both sides.

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

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

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

Subtract $3x$ from both sides.

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

Divide both sides by $2$.

$$\frac{2x^{2}-3x}{2}=-\frac{1}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{3}{2}x=-\frac{1}{2}$$

Divide $-\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{4}$. Then add the square of $-\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add $-\frac{1}{2}$ to $\frac{9}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor $x^{2}-\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{1}{4}$$ $$x-\frac{3}{4}=-\frac{1}{4}$$

Add $\frac{3}{4}$ to both sides of the equation.

$$x=1$$ $$x=\frac{1}{2}$$