Precalculus Examples

$x^{6} - 6 x^{3} + 9 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $x^{6}$ as ${(x^{3})}^{2}$ .

${(x^{3})}^{2} - 6 x^{3} + 9 = 0$

Step 1.2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$u^{2} - 6 u + 9 = 0$

Step 1.3

Factor using the perfect square rule.

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Step 1.3.1

Rewrite $9$ as $3^{2}$ .

$u^{2} - 6 u + 3^{2} = 0$

Step 1.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Step 1.3.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 3 + 3^{2} = 0$

Step 1.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

${(u - 3)}^{2} = 0$

Step 1.4

Replace all occurrences of $u$ with $x^{3}$ .

${(x^{3} - 3)}^{2} = 0$

Step 2

Set the $x^{3} - 3$ equal to $0$ .

$x^{3} - 3 = 0$

Step 3

Solve for $x$ .

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Step 3.1

Add $3$ to both sides of the equation.

$x^{3} = 3$

Step 3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt[3]{3}$

Decimal Form:

$x = 1.44224957 \dots$