What is expand 1/3 (1=3A+B+(-9)C+D) 1/3 ?

The solution to expand 1/3 (1=3A+B+(-9)C+D) 1/3 is 1/9 =3A+B/9+D/9-C

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expand 1/3 (1=3A+B+(-9)C+D) 1/3

expand

\frac{1}{3} (1 = 3 A + B + (- 9) C + D) \frac{1}{3}

\frac{1}{9} = 3 A + \frac{B}{9} + \frac{D}{9} - C

Solution steps

\frac{1}{3} (1 = 3 A + B + (- 9) C + D) \frac{1}{3}

= \frac{1}{3} \cdot \frac{1}{3} (1 \cdot 3 = A + B + (- 9) C + D)

Distribute parentheses

\frac{1}{3} \cdot \frac{1}{3} \cdot 1 = 3 A + \frac{1}{3} \cdot \frac{1}{3} B + \frac{1}{3} \cdot \frac{1}{3} (- 9) C + \frac{1}{3} \cdot \frac{1}{3} D

1 \cdot \frac{1}{3} \cdot \frac{1}{3} = 3 A + \frac{1}{3} \cdot \frac{1}{3} B + (- 9) \frac{1}{3} \cdot \frac{1}{3} C + \frac{1}{3} \cdot \frac{1}{3} D

Simplify

1 \cdot \frac{1}{3} \cdot \frac{1}{3} = 3 A + \frac{1}{3} \cdot \frac{1}{3} B + (- 9) \frac{1}{3} \cdot \frac{1}{3} C + \frac{1}{3} \cdot \frac{1}{3} D : \frac{1}{9} = 3 A + \frac{B}{9} + \frac{D}{9} - C

\frac{1}{9} = 3 A + \frac{B}{9} + \frac{D}{9} - C

(3 + 2 z i)^{2}

\frac{1}{81} (sin (9 (t - x))) + 9 x cos ((t - x) 9)

- \frac{n ( d ^{2} v + e d v + d v f - d b + e v f - e b )}{m c ( d + e )}

(1.6707100000000 \dots + 0.8976700000000 \dots \cdot i)^{4}

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

What is expand 1/3 (1=3A+B+(-9)C+D) 1/3 ?
The solution to expand 1/3 (1=3A+B+(-9)C+D) 1/3 is 1/9 =3A+B/9+D/9-C