The correct option is C a^2+4ab+16b^2/a+7b Given, a^3 64b^3/a^2+3ab 28b^2 = (a)^3 (4b)^3a^2+7ab 4ab 28b^2 = (a 4b)[(a)^2+(a)(4b)+(4b)^2]/a( ...

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Properties of Iota

a3-64b3/a2+3a...

Question

$\frac{a^{3} - 64 b^{3}}{a^{2} + 3 a b - 28 b^{2}} = ?$

\frac{a - 4 b}{a + 3 b}

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\frac{a^{2} + 4 a b + 16 b^{2}}{a - 4 b}

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\frac{a^{2} - 2 a b + 3 b^{2}}{a + 2 b}

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\frac{a^{2} + 4 a b + 16 b^{2}}{a + 7 b}

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Solution

The correct option is C $\frac{a^{2} + 4 a b + 16 b^{2}}{a + 7 b}$
Given, $\frac{a^{3} - 64 b^{3}}{a^{2} + 3 a b - 28 b^{2}}$

$= \frac{(a)^{3} - (4 b)^{3}}{a^{2} + 7 a b - 4 a b - 28 b^{2}}$
$= \frac{(a - 4 b) [(a)^{2} + (a) (4 b) + (4 b)^{2}]}{a (a + 7 b) - 4 b (a + 7 b)}$
$= \frac{(a - 4 b) (a^{2} + 4 a b + 16 b^{2})}{(a - 4 b) (a + 7 b)}$
$= \frac{a^{2} + 4 a b + 16 b^{2}}{a + 7 b}$
Option D is correct.

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1

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27 a^{3} + \frac{1}{64 b^{3}} + \frac{27 a^{2}}{4 b} + \frac{9 a}{16 b^{2}}

Simplify :
${(a - b)}^{2} + 4 a b$

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