Evaluate 1/9-(1/9)^2+(1/9)^3

$\frac{1}{9} - {(\frac{1}{9})}^{2} + {(\frac{1}{9})}^{3}$

Simplify each term.

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Apply the product rule to $\frac{1}{9}$ .

$\frac{1}{9} - \frac{1^{2}}{9^{2}} + {(\frac{1}{9})}^{3}$

One to any power is one.

$\frac{1}{9} - \frac{1}{9^{2}} + {(\frac{1}{9})}^{3}$

Raise $9$ to the power of $2$ .

$\frac{1}{9} - \frac{1}{81} + {(\frac{1}{9})}^{3}$

Apply the product rule to $\frac{1}{9}$ .

$\frac{1}{9} - \frac{1}{81} + \frac{1^{3}}{9^{3}}$

One to any power is one.

$\frac{1}{9} - \frac{1}{81} + \frac{1}{9^{3}}$

Raise $9$ to the power of $3$ .

$\frac{1}{9} - \frac{1}{81} + \frac{1}{729}$

To write $\frac{1}{9}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\frac{1}{9} \cdot \frac{9}{9} - \frac{1}{81} + \frac{1}{729}$

Write each expression with a common denominator of $81$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{9}$ and $\frac{9}{9}$ .

$\frac{9}{9 \cdot 9} - \frac{1}{81} + \frac{1}{729}$

Multiply $9$ by $9$ .

$\frac{9}{81} - \frac{1}{81} + \frac{1}{729}$

Combine the numerators over the common denominator.

$\frac{9 - 1}{81} + \frac{1}{729}$

Subtract $1$ from $9$ .

$\frac{8}{81} + \frac{1}{729}$

To write $\frac{8}{81}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\frac{8}{81} \cdot \frac{9}{9} + \frac{1}{729}$

Write each expression with a common denominator of $729$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{8}{81}$ and $\frac{9}{9}$ .

$\frac{8 \cdot 9}{81 \cdot 9} + \frac{1}{729}$

Multiply $81$ by $9$ .

$\frac{8 \cdot 9}{729} + \frac{1}{729}$

Combine the numerators over the common denominator.

$\frac{8 \cdot 9 + 1}{729}$

Simplify the numerator.

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Multiply $8$ by $9$ .

$\frac{72 + 1}{729}$

Add $72$ and $1$ .

$\frac{73}{729}$

The result can be shown in multiple forms.

Exact Form:

$\frac{73}{729}$

Decimal Form:

$0.10013717 \dots$

Evaluate 1/9-(1/9)^2+(1/9)^3

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