What is 4/(k^2+14kn+49n^2)*(k^2-49n^2)/(4k-28n) ?

The solution to 4/(k^2+14kn+49n^2)*(k^2-49n^2)/(4k-28n) is 1/(k+7n)

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simplify 4/(k^2+14kn+49n^2)*(k^2-49n^2)/(4k-28n)

simplify

\frac{4}{k ^{2} + 14 kn + 49 n ^{2}} \cdot \frac{k ^{2} - 49 n ^{2}}{4 k - 28 n}

\frac{1}{k + 7 n}

Solution steps

\frac{4}{k ^{2} + 14 kn + 49 n ^{2}} \cdot \frac{k ^{2} - 49 n ^{2}}{4 k - 28 n}

Cancel

\frac{k ^{2} - 49 n ^{2}}{4 k - 28 n} : \frac{k + 7 n}{4}

= \frac{4}{k ^{2} + 14 kn + 49 n ^{2}} \cdot \frac{k + 7 n}{4}

Cross-cancel common factor:

4

= \frac{k + 7 n}{k ^{2} + 14 kn + 49 n ^{2}}

Factor

k^{2} + 14 kn + 49 n^{2} : (k + 7 n)^{2}

= \frac{k + 7 n}{( k + 7 n ) ^{2}}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(k + 7 n)^{2} = (k + 7 n) (k + 7 n)

= \frac{k + 7 n}{( k + 7 n ) ( k + 7 n )}

Cancel the common factor:

k + 7 n

= \frac{1}{k + 7 n}

(- 729)^{\frac{1}{3}}

f (k) = \frac{3 k ^{2} - 84 k + 588}{784 - 4 k ^{2}}

x^{2} - \frac{7}{6} x + \frac{1}{3} = 0

63 = x^{2} - 2 x

\frac{0}{5}

What is 4/(k^2+14kn+49n^2)*(k^2-49n^2)/(4k-28n) ?
The solution to 4/(k^2+14kn+49n^2)*(k^2-49n^2)/(4k-28n) is 1/(k+7n)