What is (k+3)/k*(k^3-6k^2+9k)/(k^2-9) ?

The solution to (k+3)/k*(k^3-6k^2+9k)/(k^2-9) is k-3

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simplify (k+3)/k*(k^3-6k^2+9k)/(k^2-9)

simplify

\frac{k + 3}{k} \cdot \frac{k ^{3} - 6 k ^{2} + 9 k}{k ^{2} - 9}

k - 3

Solution steps

\frac{k + 3}{k} \cdot \frac{k ^{3} - 6 k ^{2} + 9 k}{k ^{2} - 9}

Cancel

\frac{k ^{3} - 6 k ^{2} + 9 k}{k ^{2} - 9} : \frac{k ( k - 3 )}{k + 3}

= \frac{k + 3}{k} \cdot \frac{k ( k - 3 )}{k + 3}

Cross-cancel common factor:

k + 3

= \frac{k ( k - 3 )}{k}

Cancel the common factor:

k

= k - 3

3 \cdot (2 x + 3 y)^{2} - 24 (2 x + 3 y) + 48

(2 x - 3)^{3} - 8 x + 12

b (2 a + a^{2} - 2 - a)

(a + b + c)^{2} + (a - b - c)^{2} + (b - c - a)^{2} + (c - b - a)^{2}

a^{10} a^{8} \cdot b^{3} b b^{2} \cdot c^{2}

What is (k+3)/k*(k^3-6k^2+9k)/(k^2-9) ?
The solution to (k+3)/k*(k^3-6k^2+9k)/(k^2-9) is k-3