Question

© Factorise :
$\frac{1}{27}(2x + 5y) {}^{3} + ( - \frac{5 }{3} y + \frac{3}{4} z) {}^{3} + ( - \frac{3}{4} z - \frac{2}{3} x) {}^{3}$
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Answer 1

Answer:

Hii mate

your solution ✌✌✌

Answer 2

SOLUTION

$\frac{1}{27}(2x + 5y) {}^{3} + ( \frac{ - 5}{3}y + \frac{3}{4} z) {}^{3} + ( \frac{ - 3}{4} z - \frac{2}{3} x) {}^{3} \\\\ = > [ \frac{1}{ 3} (2x + 5y)] {}^{3} + ( \frac{ - 5}{3} y + \frac{3}{4} z) {}^{3} + ( \frac{ - 3}{4} z - \frac{2}{3} x) {}^{3} \\ \\ = > ( \frac{2}{3} x + \frac{5}{3} y) {}^{3} + ( \frac{ - 5}{3} y + \frac{3}{4} z) {}^{3} + ( \frac{ - 3}{4} z - \frac{2}{3} x) {}^{3}$

We know that, a^3+ b^3+c^3= 3abc, if a+b+c=0

Here, a+b+c =

$= > ( \frac{2}{3} x + \frac{5}{3} y) + ( \frac{ - 5}{3}y + \frac{3}{4} z) + ( \frac{ - 3}{4} z - \frac{2}{3} x) </strong><strong>\</strong><strong>\</strong><strong>\\ = > \frac{2}{3} x + \frac{5}{3} y - \frac{5}{3} y + \frac{3}{4} z - \frac{3}{4} z - \frac{2}{3} x = 0 \\ so \\ = > ( \frac{2}{3} x + \frac{5}{3} y) {}^{3} + ( \frac{ - 5}{3} y + \frac{3}{4} z) {}^{3} + ( \frac{ - 3}{4} z - \frac{2}{3} x) {}^{3} = 3( \frac{2}{3} x + \frac{5}{3} y)( \frac{ - 5}{3}y + \frac{3}{4} z)( \frac{ - 3}{4} z - \frac{2}{3} x)$

hope it helps ☺️