What is (1-a^2)/(4a+8b)\times (a^2+4ab+4b^2)/(3-3a) ?

The solution to (1-a^2)/(4a+8b)\times (a^2+4ab+4b^2)/(3-3a) is ((1+a)(a+2b))/(12)

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simplify (1-a^2)/(4a+8b)\times (a^2+4ab+4b^2)/(3-3a)

simplify

\frac{1 - a ^{2}}{4 a + 8 b} \cdot \frac{a ^{2} + 4 ab + 4 b ^{2}}{3 - 3 a}

\frac{( 1 + a ) ( a + 2 b )}{12}

Solution steps

\frac{1 - a ^{2}}{4 a + 8 b} \cdot \frac{a ^{2} + 4 ab + 4 b ^{2}}{3 - 3 a}

Apply the fraction rule:

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

= \frac{( 1 - a ^{2} ) ( a ^{2} + 4 ab + 4 b ^{2} )}{( 4 a + 8 b ) ( 3 - 3 a )}

Factor

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

= \frac{( 1 + a ) ( 1 - a ) ( a ^{2} + 4 ab + 4 b ^{2} )}{( 4 a + 8 b ) ( 3 - 3 a )}

Factor

3 - 3 a : 3 (1 - a)

= \frac{( 1 + a ) ( 1 - a ) ( a ^{2} + 4 ab + 4 b ^{2} )}{( 4 a + 8 b ) \cdot 3 ( 1 - a )}

Cancel the common factor:

1 - a

= \frac{( 1 + a ) ( a ^{2} + 4 ab + 4 b ^{2} )}{( 4 a + 8 b ) \cdot 3}

Factor

a^{2} + 4 ab + 4 b^{2} : (a + 2 b)^{2}

= \frac{( 1 + a ) ( a + 2 b ) ^{2}}{( 4 a + 8 b ) \cdot 3}

Factor

4 a + 8 b : 4 (a + 2 b)

= \frac{( 1 + a ) ( a + 2 b ) ^{2}}{4 ( a + 2 b ) \cdot 3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(a + 2 b)^{2} = (a + 2 b) (a + 2 b)

= \frac{( 1 + a ) ( a + 2 b ) ( a + 2 b )}{4 ( a + 2 b ) \cdot 3}

Cancel the common factor:

a + 2 b

= \frac{( 1 + a ) ( a + 2 b )}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{( 1 + a ) ( a + 2 b )}{12}

0 = 200 - 32 t

5 + 6 x = 2

\frac{z + 4}{3} = \frac{z - 1}{2}

- ∣ 2 x - 1 ∣ \leq - 13

x - 9 \geq x^{2} + 2 x - 3

What is (1-a^2)/(4a+8b)\times (a^2+4ab+4b^2)/(3-3a) ?
The solution to (1-a^2)/(4a+8b)\times (a^2+4ab+4b^2)/(3-3a) is ((1+a)(a+2b))/(12)