What is 19683((a^3-243ab)/(729)+c) ?

The solution to 19683((a^3-243ab)/(729)+c) is 19683c+27a^3-6561ab

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simplify 19683((a^3-243ab)/(729)+c)

simplify

19683 (\frac{a ^{3} - 243 ab}{729} + c)

19683 c + 27 a^{3} - 6561 ab

Solution steps

19683 (\frac{a ^{3} - 243 ab}{729} + c)

Join

\frac{a ^{3} - 243 ab}{729} + c : \frac{c \cdot 729 + a ^{3} - 243 ab}{729}

= 19683 \cdot \frac{c \cdot 729 + a ^{3} - 243 ab}{729}

= 19683 \cdot \frac{729 c + a ^{3} - 243 ab}{729}

Apply the fraction rule:

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

= \frac{19683 ( 729 c + a ^{3} - 243 ab )}{729}

Divide the numbers:

\frac{19683}{729} = 27

= 27 (729 c + a^{3} - 243 ab)

Expand

27 (729 c + a^{3} - 243 ab) : 19683 c + 27 a^{3} - 6561 ab

= 19683 c + 27 a^{3} - 6561 ab

\frac{1}{x ( x + 4 )}

3 (7 - 1)

\frac{( 5 y + 8 ) ( 25 y + 40 )}{128 y}

\frac{2 s + 3}{s ( s + 1 ) ( 2 s ^{2} + 3 s - 5 )}

(2 - 4 x)^{2}

What is 19683((a^3-243ab)/(729)+c) ?
The solution to 19683((a^3-243ab)/(729)+c) is 19683c+27a^3-6561ab