Solve (27n-45)/(-9n^(2) 27n-20) ⋅ (30n^(3)-40n^(2))/(

Question

Simplify the expression

- \frac{90 n ^{4} - 270 n ^{3} + 200 n ^{2}}{243 n ^{4} - 1944 n ^{3} + 20 n - 160}

Evaluate

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

Multiply

More Steps

\frac{27 n - 45}{- 243 n ^{3} - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

- \frac{27 n - 45}{243 n ^{3} + 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

Rewrite the expression

- \frac{9 ( 3 n - 5 )}{243 n ^{3} + 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

Rewrite the expression

- \frac{9 ( 3 n - 5 )}{243 n ^{3} + 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 ( n - 8 )}

Cancel out the common factor 9

- \frac{3 n - 5}{243 n ^{3} + 20} \times \frac{30 n ^{3} - 40 n ^{2}}{n - 8}

Multiply the terms

- \frac{( 3 n - 5 ) ( 30 n ^{3} - 40 n ^{2} )}{( 243 n ^{3} + 20 ) ( n - 8 )}

Multiply the terms

More Steps

- \frac{90 n ^{4} - 270 n ^{3} + 200 n ^{2}}{( 243 n ^{3} + 20 ) ( n - 8 )}

Solution

More Steps

- \frac{90 n ^{4} - 270 n ^{3} + 200 n ^{2}}{243 n ^{4} - 1944 n ^{3} + 20 n - 160}

Show Solution

n = - \frac{3 60}{9}, n = 8

Evaluate

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

To find the excluded values,set the denominators equal to 0

- 9 n^{2} \times 27 n - 20 = 0 9 n - 72 = 0

Solve the equations

More Steps

n = - \frac{3 60}{9} 9 n - 72 = 0

Solve the equations

More Steps

n = - \frac{3 60}{9} n = 8

Solution

n = - \frac{3 60}{9}, n = 8

Show Solution

n_{1} = 0, n_{2} = \frac{4}{3}, n_{3} = \frac{5}{3}

Alternative Form

n_{1} = 0, n_{2} = 1. \dot{3}, n_{3} = 1. \dot{6}

Evaluate

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72}

To find the roots of the expression,set the expression equal to 0

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72} = 0

Find the domain

More Steps

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72} = 0, n \in (- \infty, - \frac{3 60}{9}) \cup (- \frac{3 60}{9}, 8) \cup (8, + \infty)

Calculate

\frac{27 n - 45}{- 9 n ^{2} \times 27 n - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72} = 0

Multiply

More Steps

\frac{27 n - 45}{- 243 n ^{3} - 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72} = 0

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

- \frac{27 n - 45}{243 n ^{3} + 20} \times \frac{30 n ^{3} - 40 n ^{2}}{9 n - 72} = 0

Multiply the terms

More Steps

- \frac{( 3 n - 5 ) ( 30 n ^{3} - 40 n ^{2} )}{( 243 n ^{3} + 20 ) ( n - 8 )} = 0

Rewrite the expression

\frac{( - 3 n + 5 ) ( 30 n ^{3} - 40 n ^{2} )}{( 243 n ^{3} + 20 ) ( n - 8 )} = 0

Cross multiply

(- 3 n + 5) (30 n^{3} - 40 n^{2}) = (243 n^{3} + 20) (n - 8) \times 0

Simplify the equation

(- 3 n + 5) (30 n^{3} - 40 n^{2}) = 0

Change the sign

(3 n - 5) (30 n^{3} - 40 n^{2}) = 0

Separate the equation into 2 possible cases

3 n - 5 = 0 30 n^{3} - 40 n^{2} = 0

Solve the equation

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n = \frac{5}{3} 30 n^{3} - 40 n^{2} = 0

Solve the equation

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n = \frac{5}{3} n = 0 n = \frac{4}{3}

Check if the solution is in the defined range

n = \frac{5}{3} n = 0 n = \frac{4}{3}, n \in (- \infty, - \frac{3 60}{9}) \cup (- \frac{3 60}{9}, 8) \cup (8, + \infty)

Find the intersection of the solution and the defined range

n = \frac{5}{3} n = 0 n = \frac{4}{3}

Solution

n_{1} = 0, n_{2} = \frac{4}{3}, n_{3} = \frac{5}{3}

Alternative Form

n_{1} = 0, n_{2} = 1. \dot{3}, n_{3} = 1. \dot{6}

Show Solution