What is (x^3-27)/(a^3-1)*(a^2+a+1)/(x^2+3x+9) ?

The solution to (x^3-27)/(a^3-1)*(a^2+a+1)/(x^2+3x+9) is (x-3)/(a-1)

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simplify (x^3-27)/(a^3-1)*(a^2+a+1)/(x^2+3x+9)

simplify

\frac{x ^{3} - 27}{a ^{3} - 1} \cdot \frac{a ^{2} + a + 1}{x ^{2} + 3 x + 9}

\frac{x - 3}{a - 1}

Solution steps

\frac{x ^{3} - 27}{a ^{3} - 1} \cdot \frac{a ^{2} + a + 1}{x ^{2} + 3 x + 9}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( x ^{3} - 27 ) ( a ^{2} + a + 1 )}{( a ^{3} - 1 ) ( x ^{2} + 3 x + 9 )}

Factor

x^{3} - 27 : (x - 3) (x^{2} + 3 x + 9)

= \frac{( x - 3 ) ( x ^{2} + 3 x + 9 ) ( a ^{2} + a + 1 )}{( a ^{3} - 1 ) ( x ^{2} + 3 x + 9 )}

Cancel the common factor:

x^{2} + 3 x + 9

= \frac{( x - 3 ) ( a ^{2} + a + 1 )}{a ^{3} - 1}

Factor

a^{3} - 1 : (a - 1) (a^{2} + a + 1)

= \frac{( x - 3 ) ( a ^{2} + a + 1 )}{( a - 1 ) ( a ^{2} + a + 1 )}

Cancel the common factor:

a^{2} + a + 1

= \frac{x - 3}{a - 1}

\frac{x ^{2} + 2 x - 8}{x - 2}

x (2 x^{- \frac{1}{4}})^{4}

- 15 + 4

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

- 3 + 16

What is (x^3-27)/(a^3-1)*(a^2+a+1)/(x^2+3x+9) ?
The solution to (x^3-27)/(a^3-1)*(a^2+a+1)/(x^2+3x+9) is (x-3)/(a-1)