What is expand ((3A-3C)/8)^2 ?

The solution to expand ((3A-3C)/8)^2 is (9A^2)/(64)-(9AC)/(32)+(9C^2)/(64)

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expand ((3A-3C)/8)^2

expand

(\frac{3 A - 3 C}{8})^{2}

\frac{9 A ^{2}}{64} - \frac{9 A C}{32} + \frac{9 C ^{2}}{64}

Solution steps

(\frac{3 A - 3 C}{8})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 A - 3 C ) ^{2}}{8 ^{2}}

(3 A - 3 C)^{2} = 9 A^{2} - 18 A C + 9 C^{2}

= \frac{9 A ^{2} - 18 A C + 9 C ^{2}}{8 ^{2}}

8^{2} = 64

= \frac{9 A ^{2} - 18 A C + 9 C ^{2}}{64}

Apply the fraction rule:

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

\frac{9 A ^{2} - 18 A C + 9 C ^{2}}{64} = \frac{9 A ^{2}}{64} - \frac{18 A C}{64} + \frac{9 C ^{2}}{64}

= \frac{9 A ^{2}}{64} - \frac{18 A C}{64} + \frac{9 C ^{2}}{64}

Cancel

\frac{18 A C}{64} : \frac{9 A C}{32}

= \frac{9 A ^{2}}{64} - \frac{9 A C}{32} + \frac{9 C ^{2}}{64}

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

4 x + x

(3 x + 3 i y)^{2}

\frac{180}{1500} + \frac{1}{3} (\frac{180}{1500} - 0.06)

lo g_{3} (\frac{x}{y ^{3} z ^{2}})

What is expand ((3A-3C)/8)^2 ?
The solution to expand ((3A-3C)/8)^2 is (9A^2)/(64)-(9AC)/(32)+(9C^2)/(64)