What is (n^4-81)/(n^2-6n+9) ?

The solution to (n^4-81)/(n^2-6n+9) is ((n^2+9)(n+3))/(n-3)

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simplify (n^4-81)/(n^2-6n+9)

simplify

\frac{n ^{4} - 81}{n ^{2} - 6 n + 9}

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

Solution steps

\frac{n ^{4} - 81}{n ^{2} - 6 n + 9}

Factor

n^{4} - 81 : (n^{2} + 9) (n + 3) (n - 3)

= \frac{( n ^{2} + 9 ) ( n + 3 ) ( n - 3 )}{n ^{2} - 6 n + 9}

Factor

n^{2} - 6 n + 9 : (n - 3)^{2}

= \frac{( n ^{2} + 9 ) ( n + 3 ) ( n - 3 )}{( n - 3 ) ^{2}}

Apply exponent rule:

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

(n - 3)^{2} = (n - 3) (n - 3)

= \frac{( n ^{2} + 9 ) ( n + 3 ) ( n - 3 )}{( n - 3 ) ( n - 3 )}

Cancel the common factor:

n - 3

= \frac{( n ^{2} + 9 ) ( n + 3 )}{n - 3}

16 a^{6} b^{7} c - 12 a^{5} b^{2} c^{3} + 20 a^{3} b^{10}

6 \div \frac{3}{8}

2 7^{\frac{2}{6}}

(- 1) \cdot (- 1)

2 x^{4} + x^{3} - 19 x^{2} - 9 x + 9

What is (n^4-81)/(n^2-6n+9) ?
The solution to (n^4-81)/(n^2-6n+9) is ((n^2+9)(n+3))/(n-3)