Question

Simplify:
$\frac{ {a}^{3 } + \: 27 {b}^{3} }{a - 3b} \times \frac{ {a}^{2} - 9 {b}^{2} }{ {a}^{2} - 3ab + 9 {b}^{2} } \times \frac{1}{2a - 6b}$
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Answer 1

Answer:

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Answer 2

Simplify:

Step-by-step explanation:

$\ Given \ value:\\\\\frac{a^3+27b^3}{a-3b} \times \frac{a^2-9b^2}{a^2-3ab+9b^2} \times\frac{1}{2a-6b}\\\\\ Solution: \\\\\frac{a^3+27b^3}{a-3b} \times \frac{a^2-9b^2}{a^2-3ab+9b^2} \times\frac{1}{2a-6b}\\\\\rightarrow \frac{(a^3+(3b)^3)}{a-3b} \times \frac{(a)^2-(3b)^2}{a^2-3ab+9b^2} \times\frac{1}{2(a-3b)}\\\\\rightarrow \frac{(a+3b)(a^2-3ab+9b^2)}{a-3b} \times \frac{(a-3b)(a+3b)}{a^2-3ab+9b^2} \times\frac{1}{2(a-3b)}$

$\rightarrow \frac{(a+3b)}{a-3b} \times \frac{(a+3b)}{1} \times\frac{1}{2}\\\\\rightarrow \frac{(a+3b)^2}{2(a-3b)}$

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