What is (z^2+2bz+b^2)/(3z+3b) ?

The solution to (z^2+2bz+b^2)/(3z+3b) is (z+b)/3

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simplify (z^2+2bz+b^2)/(3z+3b)

simplify

\frac{z ^{2} + 2 b z + b ^{2}}{3 z + 3 b}

\frac{z + b}{3}

Solution steps

\frac{z ^{2} + 2 b z + b ^{2}}{3 z + 3 b}

Apply Perfect Square Formula:

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

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

= \frac{( z + b ) ^{2}}{3 z + 3 b}

Factor out common term

3

= \frac{( z + b ) ^{2}}{3 ( z + b )}

Apply exponent rule:

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

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

= \frac{( z + b ) ( z + b )}{3 ( z + b )}

Cancel the common factor:

z + b

= \frac{z + b}{3}

y^{2} - 5 y - 84

\frac{1}{4} y^{6} + \frac{7}{8} + \frac{1}{64 y ^{6}}

\frac{y ^{7}}{y ^{2}}

(- 16)^{\frac{3}{4}}

2 (x + 1) = 3 + 2 x

What is (z^2+2bz+b^2)/(3z+3b) ?
The solution to (z^2+2bz+b^2)/(3z+3b) is (z+b)/3