Evaluate Using the Given Value Evaluate (m^2-3n^2k)/(m^3+2) ; if k=1/3 , n=-1 , m=4

Evaluate $\frac{m^{2} - 3 n^{2} k}{m^{3} + 2}$ ; if $k = \frac{1}{3}$ , $n = - 1$ , $m = 4$

Step 1

Replace the variable $k$ with $\frac{1}{3}$ in the expression.

$\frac{m^{2} - 3 n^{2} \frac{1}{3}}{m^{3} + 2}$

Step 2

Replace the variable $n$ with $- 1$ in the expression.

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

Step 3

Replace the variable $m$ with $4$ in the expression.

$\frac{{(4)}^{2} - 3 {(- 1)}^{2} \frac{1}{3}}{{(4)}^{3} + 2}$

Step 4

Simplify the numerator.

Raise $4$ to the power of $2$ .

$\frac{16 - 3 {(- 1)}^{2} \frac{1}{3}}{4^{3} + 2}$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3 {(- 1)}^{2}$ .

$\frac{16 + 3 (- {(- 1)}^{2}) \frac{1}{3}}{4^{3} + 2}$

Cancel the common factor.

$\frac{16 + 3 (- {(- 1)}^{2}) \frac{1}{3}}{4^{3} + 2}$

Rewrite the expression.

$\frac{16 - {(- 1)}^{2}}{4^{3} + 2}$

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

Raise $- 1$ to the power of $1$ .

$\frac{16 + {(- 1)}^{1} {(- 1)}^{2}}{4^{3} + 2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{16 + {(- 1)}^{1 + 2}}{4^{3} + 2}$

Add $1$ and $2$ .

$\frac{16 + {(- 1)}^{3}}{4^{3} + 2}$

Raise $- 1$ to the power of $3$ .

$\frac{16 - 1}{4^{3} + 2}$

Subtract $1$ from $16$ .

$\frac{15}{4^{3} + 2}$

Step 5

Simplify the denominator.

Raise $4$ to the power of $3$ .

$\frac{15}{64 + 2}$

Add $64$ and $2$ .

$\frac{15}{66}$

Step 6

Cancel the common factor of $15$ and $66$ .

Factor $3$ out of $15$ .

$\frac{3 (5)}{66}$

Cancel the common factors.

Factor $3$ out of $66$ .

$\frac{3 \cdot 5}{3 \cdot 22}$

Cancel the common factor.

$\frac{3 \cdot 5}{3 \cdot 22}$

Rewrite the expression.

$\frac{5}{22}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{5}{22}$

Decimal Form:

$0.2\overline{27}$

Evaluate Using the Given Value Evaluate (m^2-3n^2k)/(m^3+2) ; if k=1/3 , n=-1 , m=4

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